Electricity and Magnetism Practice Questions — JAMB UTME — Page 2

Question 21Question →

Match each electromagnetic device or component listed on the left with its corresponding function or operating principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Moving coil galvanometer matches with detecting/measuring small currents via torque; Electric motor matches with converting electrical energy to mechanical energy; Split-ring commutator matches with reversing current every half-cycle to maintain continuous rotation; Soft iron core matches with concentrating magnetic flux to create a radial magnetic field.

Each electromagnetic device or component relies on magnetic forces: moving coil galvanometers convert current to proportional coil torque; electric motors convert electrical energy into continuous mechanical rotation; split-ring commutators reverse current every half-turn to keep torque unidirectionally directed; soft iron cores concentrate magnetic flux to establish strong radial fields.

Step-by-Step Solution

1

Identify the primary operational application of the moving coil galvanometer.

It detects small currents using the torque

T = NIAB \sin\theta

exerted on a current-carrying coil in a magnetic field.

This establishes the link between galvanic deflection and electrical current measurement.

2

Identify the energy transformation principle of an electric motor.

The motor converts input electrical power into mechanical torque via magnetic force

F = BIL\sin\theta

.

This connects the motor to its fundamental mechanical output function.

3

Analyze the mechanical role of a split-ring commutator in DC devices.

It alternates current flow directions in the loop every

180^\circ

.

Without reversals, the coil would oscillate around equilibrium instead of continuously rotating.

4

Determine the ferromagnetic enhancement provided by a soft iron core.

It increases field strength and maintains a radial field orientation.

High magnetic permeability concentrates magnetic field lines effectively.

Key Concept

Operating principles and structural functions of electromagnetic devices based on magnetic torque and forces

Question 22Question →

A flat circular coil consisting of $50$ turns and enclosing an area of $0.02\text{ m}^2$ is placed in a uniform magnetic field directed vertically upwards. The magnitude of the magnetic field decreases steadily from $0.5\text{ T}$ to $0.1\text{ T}$ in a time interval of $0.2\text{ s}$ . If the total electrical resistance of the coil is $5.0\text{ }\Omega$ , what is the magnitude and direction of the induced current in the coil when viewed from above?

0.4\text{ A}

in an anticlockwise direction

0.4\text{ A}

in a clockwise direction

1.0\text{ A}

in an anticlockwise direction

1.0\text{ A}

in a clockwise direction

Show answer & explanation

Answer: $0.4\text{ A}$ in an anticlockwise direction

Answer

0.4\text{ A}

in an anticlockwise direction

According to Faraday's law, the induced e.m.f. is given by

E = N A \frac{\Delta B}{\Delta t} = 50 \times 0.02 \times \frac{0.4}{0.2} = 2.0\text{ V}

. By Ohm's law, the current magnitude is

I = \frac{2.0\text{ V}}{5.0\text{ }\Omega} = 0.4\text{ A}

. By Lenz's law, because the upward magnetic field is decreasing, the coil opposes this decrease by generating an upward magnetic field. By the right-hand grip rule, an upward magnetic field corresponds to an anticlockwise current flow when viewed from above.

Step-by-Step Solution

1

Calculate the magnitude of the rate of change of magnetic field strength

\frac{\Delta B}{\Delta t} = \frac{0.5\text{ T} - 0.1\text{ T}}{0.2\text{ s}} = \frac{0.4\text{ T}}{0.2\text{ s}} = 2.0\text{ T/s}

Faraday's law depends on the rate at which the magnetic flux changes over time.

2

Calculate the magnitude of the induced electromotive force (e.m.f.)

E = N A \left(\frac{\Delta B}{\Delta t}\right) = 50 \times 0.02\text{ m}^2 \times 2.0\text{ T/s} = 2.0\text{ V}

The total induced e.m.f. is proportional to the number of turns and the enclosed area.

3

Determine the magnitude of the induced current using Ohm's law

I = \frac{E}{R} = \frac{2.0\text{ V}}{5.0\text{ }\Omega} = 0.4\text{ A}

Current equals induced voltage divided by total coil resistance.

4

Determine the direction of the induced current using Lenz's law and the right-hand rule

Anticlockwise direction when viewed from above

The upward magnetic field is decreasing, so the induced current must produce its own upward magnetic field to oppose the reduction in magnetic flux.

Key Concept

Faraday's Law and Lenz's Law of Electromagnetic Induction

Estimated Time:2m 0s

Question 23Question →

A bar magnet is moved with its north pole approaching one end of a closed circular wire coil. According to Lenz's law, which of the following correctly describes the induced magnetic polarity at that face of the coil and the direction of the induced current as viewed from the side of the magnet?

Anticlockwise current, developing a North pole

Clockwise current, developing a South pole

Clockwise current, developing a North pole

Anticlockwise current, developing a South pole

Show answer & explanation

Answer: Anticlockwise current, developing a North pole

Answer

Anticlockwise current, developing a North pole

According to Lenz's law, an induced current flows in a direction such that its magnetic field opposes the change in magnetic flux that produced it. As the North pole of a magnet approaches the coil face, the coil must oppose this motion by establishing a North pole on that face. By the right-hand grip rule, looking at a North pole corresponds to an anticlockwise flow of current.

Step-by-Step Solution

1

Apply Lenz's law to determine the induced magnetic polarity

The near face of the coil must develop a North pole to repel the approaching North pole of the bar magnet

Lenz's law states that the direction of an induced current always opposes the magnetic flux change causing it.

2

Determine the direction of induced current corresponding to a North magnetic pole

Looking directly at a North magnetic pole, the induced current flows in an anticlockwise direction

By the right-hand rule (or N-S rule for coils), an anticlockwise current produces a magnetic field directed out of the face (North polarity).

Key Concept

Lenz's Law

Estimated Time:45s

Question 24Question →

Match each physical phenomenon or calculation involving magnetic forces on the left with its corresponding rule, equation, or physical principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1 matches with Fleming's Left-Hand Rule; 2 matches with the Right-Hand Grip Rule; 3 matches with the ratio

r = \frac{mv}{qB}

; 4 matches with the parallel current interaction law

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Each electromagnetic phenomenon correctly aligns with its governing physical rule or formula: force direction on a moving charge is determined by Fleming's Left-Hand Rule, magnetic field orientation around a wire by the Right-Hand Grip Rule, circular orbital radius by balancing magnetic force with centripetal force (

r = \frac{mv}{qB}

), and force between parallel conductors by Ampere's force law.

Step-by-Step Solution

1

Identify the directional rule for magnetic force on a moving charge.

Magnetic force direction is perpendicular to both particle velocity and magnetic field, given by Fleming's Left-Hand Rule.

Fleming's Left-Hand Rule relates thrust/force (thumb), magnetic field (forefinger), and current/positive charge motion (middle finger).

2

Identify the field mapping rule for a current-carrying wire.

Concentric magnetic field lines around a straight wire are mapped using the Right-Hand Grip Rule.

Pointing the right thumb along conventional current causes fingers to curl in the direction of the magnetic field vector.

3

Derive the trajectory equation for a charge in a magnetic field.

Equating magnetic force

qvB

to centripetal force

\frac{mv^2}{r}

yields

r = \frac{mv}{qB}

.

Because the magnetic force acts as a pure centripetal force, the charge follows a circular trajectory of fixed radius

r

.

4

Identify the force law between parallel currents.

The attractive force per length is given by

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Current

I_1

sets up a magnetic field

B_1 = \frac{\mu_0 I_1}{2\pi d}

at wire 2, producing force per length

B_1 I_2

.

Key Concept

Magnetic Force and Electromagnetism Rules and Equations

Question 25Question →

A particle carrying a positive charge of $3.2 \times 10^{-19}\text{ C}$ moves with a velocity of $4.0 \times 10^6\text{ m/s}$ into a uniform magnetic field of magnetic flux density $0.50\text{ T}$ . If the velocity vector of the particle makes an angle of $30^\circ$ with the direction of the magnetic field, what is the magnitude of the magnetic force exerted on the particle?

3.2 \times 10^{-13}\text{ N}

6.4 \times 10^{-13}\text{ N}

5.5 \times 10^{-13}\text{ N}

1.6 \times 10^{-13}\text{ N}

Show answer & explanation

Answer: $3.2 \times 10^{-13}\text{ N}$

Answer

The magnitude of the magnetic force exerted on the particle is

3.2 \times 10^{-13}\text{ N}

.

The magnetic force on a charged particle moving through a magnetic field is given by

F = q v B \sin\theta

. Substituting

q = 3.2 \times 10^{-19}\text{ C}

,

v = 4.0 \times 10^6\text{ m/s}

,

B = 0.50\text{ T}

, and

\sin(30^\circ) = 0.50

yields

F = 3.2 \times 10^{-13}\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities

q = 3.2 \times 10^{-19}\text{ C}

,

v = 4.0 \times 10^6\text{ m/s}

,

B = 0.50\text{ T}

,

\theta = 30^\circ

Extract values needed for the magnetic force formula.

2

Apply the magnetic force formula for a moving charge

F = q v B \sin\theta

The force experienced by a moving point charge in a uniform magnetic field depends on charge, speed, field strength, and the angle between velocity and field.

3

Substitute the values and calculate

F = (3.2 \times 10^{-19}) \times (4.0 \times 10^6) \times 0.50 \times \sin(30^\circ) = (6.4 \times 10^{-13}) \times 0.50 = 3.2 \times 10^{-13}\text{ N}

Since

\sin(30^\circ) = 0.5

, evaluating the expression yields

3.2 \times 10^{-13}\text{ N}

.

Key Concept

Magnetic Force on a Moving Charge (

F = q v B \sin\theta

)

Question 26Question →

A coil consisting of $50$ turns is placed in a region of changing magnetic field. If the magnetic flux passing through the coil increases uniformly from $0.2\text{ Wb}$ to $0.6\text{ Wb}$ in $2.0\text{ s}$ , what is the magnitude of the induced electromotive force in the coil in volts?

Show answer & explanation

Answer: 10

Answer

The magnitude of the induced electromotive force in the coil is

10\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force

E

is proportional to the number of turns

N

and the rate of change of magnetic flux

\frac{\Delta \Phi}{\Delta t}

. Given

N = 50

,

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

, and

\Delta t = 2.0\text{ s}

, substituting these into

E = N \frac{\Delta \Phi}{\Delta t}

gives

E = 50 \times \frac{0.4}{2.0} = 10\text{ V}

.

Step-by-Step Solution

1

Determine the change in magnetic flux through the coil

\Delta \Phi = 0.6\text{ Wb} - 0.2\text{ Wb} = 0.4\text{ Wb}

Induction depends on the change in magnetic flux over time.

2

Apply Faraday's law of electromagnetic induction to solve for the induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 50 \times \frac{0.4\text{ Wb}}{2.0\text{ s}} = 10\text{ V}

Faraday's law states that the induced e.m.f. is equal to the product of the number of turns and the rate of change of flux.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 27Question →

A metallic wire has a resistance of $10.0\,\Omega$ at $0^\circ\text{C}$ and a temperature coefficient of resistance $\alpha = 4.0 \times 10^{-3}\,\text{K}^{-1}$ . The wire is uniformly stretched at constant temperature until its length is doubled while maintaining constant mass and volume. It is subsequently heated to $50^\circ\text{C}$ . What is the electric current, in Amperes, that flows through the wire when a potential difference of $120.0\,\text{V}$ is applied across its ends?

Show answer & explanation

Answer: 2.5

Answer

The electric current passing through the heated, stretched wire is 2.5 A.

When a wire of initial resistance 10.0 ohms is stretched to twice its original length, conservation of volume requires its cross-sectional area to halve, which quadruples its resistance to 40.0 ohms at 0 °C. Heating the wire by 50 K increases its resistance by a factor of (1 + 0.004 * 50) = 1.2, producing a final resistance of 48.0 ohms. Applying a 120.0 V potential difference across 48.0 ohms results in a current of 2.5 A.

Step-by-Step Solution

1

Determine resistance change due to wire stretching

R_0' = 40.0 ohms at 0 °C

Uniform stretching conserves total volume (V = A * L). Doubling length halves area, making resistance increase by a factor of 2^2 = 4.

2

Calculate resistance at 50 °C using temperature coefficient

R(50 °C) = 48.0 ohms

Resistance increases linearly with temperature: R(T) = R_0'(1 + alpha * Delta T).

3

Apply Ohm's law to find current

I = 2.5 A

Current is given by potential difference divided by total resistance at the operational temperature.

Key Concept

Resistance variation with geometric stretching and temperature coefficient of resistance

Question 28Question →

A rectangular coil of $200$ turns has dimensions $0.15\text{ m}$ by $0.10\text{ m}$ . The coil is placed with its plane perpendicular to a uniform magnetic field of $0.40\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.06\text{ s}$ , calculate the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts.

Show answer & explanation

Answer: 40

Answer

The magnitude of the average induced electromotive force is

40\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force (e.m.f.) is given by

\mathcal{E} = N \left| \frac{\Delta \Phi}{\Delta t} \right|

. Since the coil is initially perpendicular to the magnetic field

B

, the initial flux per turn is

\Phi_i = B A

. When the field is completely reversed, the final flux becomes

\Phi_f = -B A

, giving a magnitude of flux change per turn of

|\Delta \Phi| = B A - (-B A) = 2 B A

. Substituting

N = 200

,

A = 0.015\text{ m}^2

,

B = 0.40\text{ T}

, and

\Delta t = 0.06\text{ s}

gives

\mathcal{E} = 200 \times \frac{2 \times 0.40 \times 0.015}{0.06} = 40\text{ V}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the rectangular coil

A = 0.15\text{ m} \times 0.10\text{ m} = 0.015\text{ m}^2

The area is required to determine the magnetic flux passing through the coil.

2

Compute the initial magnetic flux per turn

\Phi_i = B A = 0.40 \times 0.015 = 0.006\text{ Wb}

Magnetic flux is defined as the product of magnetic field strength and area when perpendicular.

3

Calculate the change in flux per turn when the magnetic field reverses direction

|\Delta \Phi| = \Phi_i - (-\Phi_i) = 2 \Phi_i = 0.012\text{ Wb}

Reversing the field flips the direction of the flux vectors, resulting in a net change equal to twice the magnitude of the initial flux.

4

Apply Faraday's Law of Electromagnetic Induction to find the induced e.m.f.

\mathcal{E} = N \frac{|\Delta \Phi|}{\Delta t} = 200 \times \frac{0.012}{0.06} = 40\text{ V}

The magnitude of induced e.m.f. equals the total rate of change of magnetic flux linkage across all turns.

Key Concept

Faraday's Law of Electromagnetic Induction (Magnetic Flux Reversal)

Question 29Question →

A rectangular wire loop is pulled horizontally to the right out of a region containing a uniform magnetic field directed perpendicularly into the page. According to Lenz's law, what is the direction of the induced current in the loop and the direction of the resulting magnetic force acting on the loop?

Clockwise induced current; magnetic force directed to the left

Counter-clockwise induced current; magnetic force directed to the left

Clockwise induced current; magnetic force directed to the right

Counter-clockwise induced current; magnetic force directed to the right

Show answer & explanation

Answer: Clockwise induced current; magnetic force directed to the left

Answer

Clockwise induced current; magnetic force directed to the left

According to Lenz's law, an induced electric current flows in a direction such that its magnetic field opposes the change in magnetic flux that produced it. As the loop is pulled to the right, the magnetic flux pointing into the page decreases. The loop responds by inducing a current that creates additional magnetic field into the page to resist this decrease. By the right-hand rule, a current circulating clockwise produces a magnetic field into the page. Additionally, the magnetic force created on the loop must oppose the external motion, acting to the left.

Step-by-Step Solution

1

Determine the change in magnetic flux passing through the loop

As the rectangular loop is pulled to the right out of the magnetic field region, the magnetic flux directed into the page through the loop is decreasing.

Electromagnetic induction depends on the rate of change of magnetic flux.

2

Apply Lenz's law to determine the direction of the induced magnetic field and current

To oppose the decrease in inward flux, the induced current must create its own magnetic field directed into the page. By the right-hand grip rule, an inward induced field corresponds to a clockwise current flow.

Lenz's law states that the direction of an induced current always opposes the change in magnetic flux causing it.

3

Determine the direction of the net magnetic force on the loop

The induced magnetic force must oppose the mechanical motion pulling the loop to the right, so the net magnetic force acts to the left.

Lenz's law is a consequence of the conservation of energy, ensuring mechanical work must be done against electromagnetic forces.

Key Concept

Lenz's Law and Direction of Induced Current

Question 30Question →

Two point charges of $+2.0 \times 10^{-6}\text{ C}$ and $+4.0 \times 10^{-6}\text{ C}$ are placed in a vacuum at a distance of $0.3\text{ m}$ apart. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force exerted between them in Newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the electrostatic force between the charges is 0.8 N.

According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them:

F = \frac{k q_1 q_2}{r^2}

. Substituting

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

into the formula gives

F = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify known quantities from the problem statement

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

Clear identification of parameters is required for substitution into Coulomb's Law.

2

Apply Coulomb's Law formula

F = \frac{k q_1 q_2}{r^2}

Coulomb's Law quantifies the electrostatic force between two stationary point charges.

3

Substitute values and perform arithmetic calculation

F = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}

Squaring the separation distance

0.3\text{ m}

gives

0.09\text{ m}^2

, and evaluating the numerator gives

0.072\text{ N}\cdot\text{m}^2

.

Key Concept

Coulomb's Law

Question 31Question →

A metal rod of length $0.4\text{ m}$ glides at a constant speed of $5.0\text{ m s}^{-1}$ to the right along horizontal parallel conducting rails placed in a uniform magnetic field of $0.5\text{ T}$ directed vertically into the page. The rails are connected at their left end by a $2.0\ \Omega$ resistor. What is the magnitude of the external force required to maintain the uniform speed of the rod, and in which direction does the induced magnetic force act on the rod?

0.10\text{ N}

, directed to the left

0.10\text{ N}

, directed to the right

0.50\text{ N}

, directed to the left

0.50\text{ N}

, directed to the right

Show answer & explanation

Answer: $0.10\text{ N}$ , directed to the left

Answer

0.10\text{ N}

, directed to the left

The motional e.m.f. generated across the moving conductor is

\mathcal{E} = BLv = 0.5 \times 0.4 \times 5.0 = 1.0\text{ V}

. The induced current is

I = \mathcal{E}/R = 1.0 / 2.0 = 0.5\text{ A}

. The magnetic force resisting the rod's motion is

F_B = BIL = 0.5 \times 0.5 \times 0.4 = 0.10\text{ N}

. By Lenz's law, this magnetic force acts to the left, opposing the motion to the right. To maintain uniform speed, an equal external force of

0.10\text{ N}

must be applied.

Step-by-Step Solution

1

Calculate the induced electromotive force (e.m.f.) across the moving rod

\mathcal{E} = B L v = 0.5\text{ T} \times 0.4\text{ m} \times 5.0\text{ m s}^{-1} = 1.0\text{ V}

Motional e.m.f. is produced when a conductor cuts magnetic flux lines at a perpendicular velocity.

2

Calculate the induced current flowing in the circuit

I = \frac{\mathcal{E}}{R} = \frac{1.0\text{ V}}{2.0\ \Omega} = 0.5\text{ A}

Ohm's law relates the induced e.m.f. and total circuit resistance.

3

Determine the magnitude of the magnetic force acting on the current-carrying rod

F_B = B I L = 0.5\text{ T} \times 0.5\text{ A} \times 0.4\text{ m} = 0.10\text{ N}

A magnetic field exerts a force on a straight conductor carrying current.

4

Determine the direction of the magnetic force using Lenz's law and Newton's first law

The magnetic force acts to the left (opposing motion to the right). An external force of equal magnitude (

0.10\text{ N}

) to the right is required to maintain constant speed.

Lenz's law states that induced effects always oppose the change causing them (motion to the right).

Key Concept

Motional Electromotive Force, Magnetic Force on a Conductor, and Lenz's Law

Question 32Question →

When a strong bar magnet is dropped vertically through a long, hollow copper tube, its downward acceleration is equal to the acceleration due to gravity ( $g$ ) because copper is a non-magnetic material.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. The magnet falls with an acceleration less than

g

because induced eddy currents in the copper tube create an upward magnetic force that opposes the motion.

The statement is false because the relative motion between the falling magnet and the conductive copper tube induces eddy currents. According to Lenz's law, these induced currents set up a magnetic field that opposes the falling magnet's motion, creating an upward retarding force that reduces the downward acceleration to a value less than

g

.

Step-by-Step Solution

1

Identify the physical interactions as the magnet falls through the tube.

The falling magnet creates a changing magnetic flux through the surrounding copper tube.

Relative motion between a magnetic field source and a conductor produces a time-varying magnetic flux in the conductor.

2

Apply Faraday's law of electromagnetic induction.

Electromotive force (e.m.f.) and circular eddy currents are induced in the conductive copper walls.

A changing magnetic flux induces electric currents in any closed conductive path.

3

Apply Lenz's law to determine the magnetic effect of the induced eddy currents.

The induced eddy currents produce a magnetic field that opposes the downward motion of the falling magnet, generating an upward magnetic force (

F_{\text{mag}}

).

Lenz's law dictates that an induced current always flows in a direction such that its magnetic field opposes the change causing it.

4

Analyze the net force and resulting acceleration.

The net downward force is

F_{\text{net}} = mg - F_{\text{mag}} < mg

, so the downward acceleration

a = g - \frac{F_{\text{mag}}}{m} < g

.

The presence of an upward magnetic force reduces the net downward acceleration below the free-fall value of

g

.

Key Concept

Lenz's Law and Eddy Currents in Conductors

Estimated Time:1m 0s

Question 33Question →

A uniform metallic wire of resistance $R$ is stretched uniformly until its radius decreases by $20\%$ . The stretched wire is subsequently cut into two equal halves, which are then connected in parallel across a constant potential difference $V$ . What is the ratio of the total electrical power dissipated in this parallel combination to the power dissipated by the original unstretched wire under the same potential difference?

1.64

0.61

2.56

0.41

Show answer & explanation

Answer: 1.64

Answer

The ratio of the total power dissipated in the parallel combination to the original power is 1.64

When a wire of initial resistance

R

is stretched so that its radius decreases by

20\%

, its new radius is

0.8r

. Because volume is conserved (

A_1 L_1 = A_2 L_2

), reducing the cross-sectional area to

0.64A

causes the length to increase to

L/0.64

. Consequently, the resistance scales inversely with the fourth power of the radius:

R_{\text{stretched}} = R / (0.8)^4 = R / 0.4096 = 2.4414R

. Cutting this wire into two equal pieces gives two resistors of

1.2207R

each. Connecting them in parallel yields an equivalent resistance

R_{\text{eq}} = 1.2207R / 2 = 0.61035R

. Power at constant voltage is

P = V^2/R

, so the new power is

P_{\text{new}} = V^2 / (0.61035R) = 1.64 (V^2/R) = 1.64 P_{\text{orig}}

.

Step-by-Step Solution

1

Determine the new resistance of the wire after stretching

R_{\text{stretched}} = \frac{R}{(0.8)^4} = \frac{R}{0.4096} \approx 2.4414 R

Since mass and density remain constant, volume

V_{\text{vol}} = A \cdot L

is conserved. Decreasing radius to

r_2 = 0.8 r_1

reduces area to

A_2 = 0.64 A_1

and increases length to

L_2 = L_1 / 0.64

. Resistance

R = \rho L / A \propto 1/r^4

.

2

Calculate the equivalent resistance of the two equal halves connected in parallel

R_{\text{eq}} = \frac{1}{4} R_{\text{stretched}} = \frac{2.4414 R}{4} \approx 0.61035 R

Cutting the stretched wire in half gives two pieces each of resistance

R_{\text{half}} = R_{\text{stretched}} / 2

. Connecting two identical resistors in parallel yields an equivalent resistance

R_{\text{eq}} = R_{\text{half}} / 2 = R_{\text{stretched}} / 4

.

3

Calculate the ratio of power dissipated across a constant potential difference V

\frac{P_{\text{new}}}{P_{\text{orig}}} = \frac{V^2 / R_{\text{eq}}}{V^2 / R} = \frac{R}{R_{\text{eq}}} = \frac{1}{0.61035} \approx 1.64

Electrical power dissipated at constant voltage is given by

P = V^2 / R

, which means power is inversely proportional to equivalent resistance.

Key Concept

Dependence of electrical resistance on conductor geometry under volume conservation, and power dissipation in parallel circuits.

Question 34Question →

A $150\text{ V}$ (RMS) AC generator is connected across a series combination of a $12\ \Omega$ resistor, an inductor with an inductive reactance of $20\ \Omega$ , and a capacitor with a capacitive reactance of $11\ \Omega$ . What is the average electrical power consumed by this circuit?

900\text{ W}

1200\text{ W}

1500\text{ W}

1875\text{ W}

Show answer & explanation

Answer: $1200\text{ W}$

Answer

The average electrical power consumed by the circuit is

1200\text{ W}

.

To find the average real power dissipated in an AC series circuit, first determine the net reactance

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

. Next, calculate total impedance using

Z = \sqrt{R^2 + X^2} = \sqrt{12^2 + 9^2} = 15\ \Omega

. The RMS current in the circuit is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150\text{ V}}{15\ \Omega} = 10\text{ A}

. Since energy is dissipated only by resistance, the average real power is

P = I_{\text{rms}}^2 R = (10)^2 \times 12 = 1200\text{ W}

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

.

2

Calculate the total impedance of the series RLC circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance add vectorially at a right angle.

3

Determine the RMS current flowing through the circuit.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150\text{ V}}{15\ \Omega} = 10\text{ A}

Ohm's law for AC circuits states that current is the supply RMS voltage divided by total impedance.

4

Calculate the average real power dissipated by the circuit.

P_{\text{avg}} = I_{\text{rms}}^2 R = (10\text{ A})^2 \times 12\ \Omega = 1200\text{ W}

In an AC circuit, average power is only dissipated by resistive components, as ideal inductors and capacitors consume zero net real power over a complete cycle.

Key Concept

Power Dissipation in AC Series Circuits

Question 35Question →

Two equal positive point charges, each of magnitude $2.0 \times 10^{-6}\text{ C}$ , are placed $0.2\text{ m}$ apart in a vacuum. What is the magnitude of the net electric field intensity at the midpoint between the two charges? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

0\text{ N C}^{-1}

1.8 \times 10^6\text{ N C}^{-1}

3.6 \times 10^6\text{ N C}^{-1}

7.2 \times 10^6\text{ N C}^{-1}

Show answer & explanation

Answer: $0\text{ N C}^{-1}$

Answer

The net electric field intensity at the midpoint is

0\text{ N C}^{-1}

.

At the midpoint between two identical positive charges, the electric field created by each charge has the exact same magnitude because the charges and distances are equal. Because electric field lines point away from positive charges, the two field vectors at the midpoint point in directly opposite directions. Taking vector superposition gives a net electric field of zero.

Step-by-Step Solution

1

Determine the distance from each charge to the midpoint.

The midpoint distance

r = \frac{0.2\text{ m}}{2} = 0.1\text{ m}

.

Electric field calculation requires the distance from the point charge to the point of evaluation.

2

Calculate the magnitude of the electric field due to one charge.

E = \frac{k Q}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\text{ N C}^{-1}

.

Electric field intensity magnitude is given by Coulomb's field formula

E = \frac{k Q}{r^2}

.

3

Apply vector addition to find the net electric field at the midpoint.

E_{\text{net}} = E_1 - E_2 = 1.8 \times 10^6 - 1.8 \times 10^6 = 0\text{ N C}^{-1}

.

Electric field is a vector quantity. Since both charges are positive, the field vectors point away from each charge and act in opposite directions at the midpoint.

Key Concept

Vector superposition of electric fields

Question 36Question →

A coil consisting of $150$ turns is placed in a magnetic field. The magnetic flux passing through the coil decreases uniformly from $4.0 \times 10^{-3}\text{ Wb}$ to $1.0 \times 10^{-3}\text{ Wb}$ over a time interval of $0.015\text{ s}$ . What is the magnitude of the average induced electromotive force in the coil in volts?

Show answer & explanation

Answer: 30

Answer

The magnitude of the average induced electromotive force in the coil is

30\text{ V}

.

According to Faraday's law of electromagnetic induction, the magnitude of the average induced electromotive force in a coil with

N

turns is given by

E = N \frac{\Delta \Phi}{\Delta t}

. Given

N = 150

,

\Delta \Phi = 3.0 \times 10^{-3}\text{ Wb}

, and

\Delta t = 0.015\text{ s}

, the magnitude of the induced e.m.f. is

E = 150 \times \frac{3.0 \times 10^{-3}}{0.015} = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the magnitude of the change in magnetic flux through the coil

\Delta \Phi = 4.0 \times 10^{-3}\text{ Wb} - 1.0 \times 10^{-3}\text{ Wb} = 3.0 \times 10^{-3}\text{ Wb}

Faraday's law relates induced e.m.f. directly to the rate of change of magnetic flux.

2

Apply Faraday's Law of Electromagnetic Induction equation for an N-turn coil

E = N \frac{\Delta \Phi}{\Delta t}

The total induced electromotive force in a coil is proportional to the number of turns and the rate of change of flux.

3

Substitute the given numerical values to compute the magnitude of the induced e.m.f.

E = 150 \times \frac{3.0 \times 10^{-3}\text{ Wb}}{0.015\text{ s}} = 150 \times 0.2\text{ V} = 30\text{ V}

Performing clean calculation without needing a calculator.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 37Question →

A $3\text{ }\mu\text{F}$ capacitor and a $6\text{ }\mu\text{F}$ capacitor are connected in series across a direct current voltage source. What is the total equivalent capacitance of the combination, in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 2

Answer

The total equivalent capacitance of the combination is

2\text{ }\mu\text{F}

.

For capacitors connected in series, the reciprocal of the total equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. Substituting

3\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

gives

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

, which yields an equivalent capacitance of

2\text{ }\mu\text{F}

.

Step-by-Step Solution

1

State the formula for equivalent capacitance of two capacitors in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Capacitors connected in series combine reciprocally, unlike resistors connected in series.

2

Substitute the values of

C_1

and

C_2

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

Find a common denominator and add the fractions.

3

Calculate the reciprocal to determine

C_{eq}

C_{eq} = 2\text{ }\mu\text{F}

Inverting

\frac{1}{2}

yields the total equivalent capacitance.

Key Concept

Equivalent Capacitance in Series

Question 38Question →

A point charge $q_1 = +9.0 \times 10^{-9}\text{ C}$ is fixed at the origin ( $x = 0\text{ m}$ ), and a second point charge $q_2 = -4.0 \times 10^{-9}\text{ C}$ is fixed on the x-axis at $x = 0.5\text{ m}$ . At what position $x$ (in meters) along the x-axis is the net electric field intensity equal to zero?

Show answer & explanation

Answer: 1.5

Answer

The net electric field intensity is zero at

x = 1.5\text{ m}

.

The correct position is

x = 1.5\text{ m}

. At this point, the electric field from

+q_1

points in the

+x

direction with magnitude

E_1 = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-9}}{1.5^2} = 36\text{ N C}^{-1}

, and the electric field from

-q_2

points in the

-x

direction with magnitude

E_2 = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{1.0^2} = 36\text{ N C}^{-1}

. The two vectors are equal in magnitude and opposite in direction, yielding a net electric field of zero.

Step-by-Step Solution

1

Determine the physical region where electric fields can cancel

The point of zero field lies to the right of

q_2

, i.e.,

x > 0.5\text{ m}

.

Between the charges, the fields due to

+q_1

and

-q_2

point in the same direction (+x). To the left of

q_1

,

q_1

is both larger in magnitude and closer, so

E_1 > E_2

everywhere. Hence, balance can only occur to the right of the smaller magnitude charge

q_2

.

2

Set up the condition for equal electric field magnitudes

\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(x - 0.5)^2}

For the net field to be zero, the vector sum of

E_1

and

E_2

must equal zero, meaning their magnitudes must be equal.

3

Substitute values and simplify the algebraic equation

\frac{9.0 \times 10^{-9}}{x^2} = \frac{4.0 \times 10^{-9}}{(x - 0.5)^2} \implies \frac{9}{x^2} = \frac{4}{(x - 0.5)^2}

Coulomb's constant

k

and the power factor

10^{-9}

cancel from both sides.

4

Take square root on both sides to solve for x

\frac{3}{x} = \frac{2}{x - 0.5} \implies 3(x - 0.5) = 2x \implies x = 1.5\text{ m}

Taking the principal square root reduces the quadratic relation to a simple linear equation.

Key Concept

Electric Field Superposition and Zero Field Condition for Point Charges

Question 39Question →

A uniform conductor of length $50\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ is made of a material with a resistivity of $4.0 \times 10^{-7}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the conductor in ohms?

Show answer & explanation

Answer: 10

Answer

The resistance of the conductor is

10\,\Omega

.

The resistance of a uniform conductor is given by

R = \frac{\rho L}{A}

. Substituting the values

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

yields

R = \frac{(4.0 \times 10^{-7})(50)}{2.0 \times 10^{-6}} = 10\,\Omega

.

Step-by-Step Solution

1

Identify given physical quantities

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

,

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

Extracting given parameter values clearly sets up the mathematical relationship.

2

Apply the resistivity formula for resistance

R = \frac{\rho L}{A}

Resistance varies directly with length and resistivity, and inversely with cross-sectional area.

3

Perform the calculation

R = \frac{(4.0 \times 10^{-7}\,\Omega\cdot\text{m})(50\,\text{m})}{2.0 \times 10^{-6}\,\text{m}^2} = 10\,\Omega

Multiplying the numerator gives

2.0 \times 10^{-5}\,\Omega\cdot\text{m}^2

; dividing by

2.0 \times 10^{-6}\,\text{m}^2

yields

10\,\Omega

.

Key Concept

Direct computation of electrical resistance using resistivity, length, and cross-sectional area

Estimated Time:45s

Question 40Question →

Two point charges, $Q_1 = +3.0 \times 10^{-6}\text{ C}$ and $Q_2 = +4.0 \times 10^{-6}\text{ C}$ , are positioned in a vacuum at coordinates $(0\text{ m}, 3.0\text{ m})$ and $(3.0\text{ m}, 0\text{ m})$ respectively on a Cartesian plane. Taking the electrostatic constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electric field intensity at the origin $(0,0)$ ?

1.0 \times 10^3\text{ N C}^{-1}

5.0 \times 10^3\text{ N C}^{-1}

7.0 \times 10^3\text{ N C}^{-1}

1.5 \times 10^4\text{ N C}^{-1}

Show answer & explanation

Answer: $5.0 \times 10^3\text{ N C}^{-1}$

Answer

The magnitude of the net electric field intensity at the origin is

5.0 \times 10^3\text{ N C}^{-1}

.

The electric field intensity at the origin is a vector sum of the individual electric fields created by each point charge. The field due to the charge on the y-axis points downward along the y-axis with a magnitude of

3.0 \times 10^3\text{ N C}^{-1}

, while the field due to the charge on the x-axis points leftward along the x-axis with a magnitude of

4.0 \times 10^3\text{ N C}^{-1}

. Because these two fields act at right angles to each other, their vector sum magnitude is given by

\sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Step-by-Step Solution

1

Calculate the electric field intensity

E_1

at the origin due to charge

Q_1

E_1 = \frac{k |Q_1|}{r_1^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{3.0^2} = 3.0 \times 10^3\text{ N C}^{-1}

, directed along the negative y-axis.

Electric field magnitude follows Coulomb's law for field strength, and positive charges create fields directed away from themselves.

2

Calculate the electric field intensity

E_2

at the origin due to charge

Q_2

E_2 = \frac{k |Q_2|}{r_2^2} = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{3.0^2} = 4.0 \times 10^3\text{ N C}^{-1}

, directed along the negative x-axis.

The charge is located at

(3.0, 0)

on the x-axis, producing a field pointing toward the origin.

3

Calculate the net electric field vector magnitude at the origin

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Since the two component fields are perpendicular along orthogonal axes (x and y), their resultant is found using the Pythagorean theorem.

Key Concept

Vector addition of electric field intensities from multiple point charges

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