Electricity and Magnetism Practice Questions — JAMB UTME — Page 7

Question 121Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected across a load resistor of $5.0\text{ }\Omega$ . What is the terminal potential difference across the battery in volts?

Show answer & explanation

Answer: 10

Answer

The terminal potential difference across the battery is

10.0\text{ V}

.

The total resistance in the circuit is the sum of the external load resistance and the battery's internal resistance (

R_{\text{total}} = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

). The current drawn from the battery is

I = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

. The terminal potential difference available to the external load is

V = I \cdot R = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

.

Step-by-Step Solution

1

Find the total circuit resistance including internal resistance

R_{\text{total}} = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

The internal resistance of the cell acts in series with the external load resistor.

2

Determine total current in the circuit

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

By Ohm's law, current equals total electromotive force divided by total circuit resistance.

3

Compute the terminal potential difference

V = I \times R = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

Terminal voltage is the voltage drop across the load resistor, equivalent to

E - I \cdot r

.

Key Concept

Terminal Potential Difference and Internal Resistance

Question 122Question →

An electric immersion heater with a resistance of $20\,\Omega$ carries a steady current of $3\,\text{A}$ . How much electrical energy is dissipated as heat by the heater in $10\,\text{seconds}$ ?

600\,\text{J}

1200\,\text{J}

1800\,\text{J}

180\,\text{J}

Show answer & explanation

Answer: $1800\,\text{J}$

Answer

The total electrical energy dissipated as heat by the heater is

1800\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into heat energy in a resistor is given by

E = I^2 R t

. Substituting

I = 3\,\text{A}

,

R = 20\,\Omega

, and

t = 10\,\text{s}

yields

E = (3)^2 \times 20 \times 10 = 9 \times 200 = 1800\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Resistance

R = 20\,\Omega

, Current

I = 3\,\text{A}

, Time

t = 10\,\text{s}

Recognize the required values needed to compute heat energy dissipated in a resistor.

2

Select the appropriate formula for heat energy

E = I^2 R t

Joule's Law of Heating relates electrical energy, current, resistance, and time.

3

Substitute the known values into the equation and calculate

E = (3)^2 \times 20 \times 10 = 9 \times 20 \times 10 = 1800\,\text{J}

Perform arithmetic multiplication to find the final heat energy in Joules.

Key Concept

Joule's Law of Heating

Question 123Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected across a series combination of a $2.0\text{ }\Omega$ resistor and a $3.0\text{ }\Omega$ resistor. What is the potential difference across the terminals of the battery?

10.0\text{ V}

12.0\text{ V}

2.0\text{ V}

6.0\text{ V}

Show answer & explanation

Answer: $10.0\text{ V}$

Answer

10.0\text{ V}

The total resistance of the circuit is the sum of the external load resistors and the internal resistance (

2.0 + 3.0 + 1.0 = 6.0\text{ }\Omega

). The total current is

I = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

. The potential difference across the terminals of the battery is the voltage across the external load

V = I \times R_{\text{ext}} = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

(or

E - Ir = 12.0 - 2.0 = 10.0\text{ V}

).

Step-by-Step Solution

1

Calculate total external resistance in series

R_{\text{ext}} = 2.0\text{ }\Omega + 3.0\text{ }\Omega = 5.0\text{ }\Omega

Resistors in series add directly to give the total external load resistance.

2

Calculate total circuit resistance including internal resistance

R_{\text{total}} = R_{\text{ext}} + r = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

The cell's internal resistance is in series with the external circuit.

3

Determine the total current drawn from the battery

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

Applying Ohm's law to the entire circuit.

4

Calculate terminal potential difference across the battery

V = E - Ir = 12.0\text{ V} - (2.0\text{ A} \times 1.0\text{ }\Omega) = 10.0\text{ V}

Terminal voltage is the e.m.f. minus the lost volts across internal resistance (or equivalently

V = I \times R_{\text{ext}}

).

Key Concept

Terminal potential difference vs. electromotive force (e.m.f.) and lost volts

Estimated Time:1m 30s

Question 124Question →

Determine whether the following statement is true or false: When a bar magnet is placed in the Earth's magnetic field with its north pole pointing towards geographic North, the neutral points formed by the mutual cancellation of the magnet's field and the Earth's horizontal field lie along the magnet's equatorial line (broadside-on position).

True

False

Show answer & explanation

Answer: True

Answer

True. When a bar magnet has its north pole pointing North, the magnetic field along its equatorial (broadside-on) line points South, opposing the Earth's horizontal magnetic field to create neutral points.

The statement is correct because neutral points are formed at positions where the external magnetic field of a magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field component (

B_H

). When the north pole of a bar magnet points North, its field along the broadside-on (equatorial) axis is directed South (opposite to

B_H

), leading to field cancellation and the creation of two neutral points on that axis.

Step-by-Step Solution

1

Identify the direction of the Earth's horizontal magnetic field component (

B_H

).

Earth's horizontal field (

B_H

) acts along the magnetic meridian from magnetic South to magnetic North.

Neutral points can only occur where two magnetic fields act in opposite directions and have equal magnitudes.

2

Determine the direction of the bar magnet's magnetic field on its broadside-on (equatorial) axis.

Outside the magnet, magnetic flux lines travel from North to South. Along the broadside-on line, the magnet's field points southward.

Field lines curve from the north pole back into the south pole externally.

3

Compare the magnetic field directions on the broadside-on axis.

The southward field of the magnet opposes the northward Earth field (

B_H

). At points where their magnitudes match, the net magnetic field becomes zero.

Vector cancellation requires antiparallel vectors of equal magnitude.

Key Concept

Neutral Points of a Bar Magnet in Earth's Magnetic Field

Question 125Question →

An electric cell of electromotive force $E$ and internal resistance $r$ is connected in series with a fixed resistor of resistance $8.0\text{ }\Omega$ and a galvanometer of internal resistance $40.0\text{ }\Omega$ . When a shunt resistor of resistance $10.0\text{ }\Omega$ is connected in parallel across the galvanometer, the total current supplied by the cell increases by $50\%$ . What is the internal resistance $r$ of the cell in ohms?

Show answer & explanation

Answer: 48

Answer

The internal resistance of the cell is

48.0\text{ }\Omega

.

By determining the initial total circuit resistance

R_{T1} = 48.0 + r

and the post-shunt total circuit resistance

R_{T2} = 16.0 + r

, we apply Ohm's law with

I_2 = 1.5 I_1

. Equating total supply voltage

E

gives

\frac{E}{16.0 + r} = 1.5 \times \frac{E}{48.0 + r}

, which simplifies directly to

r = 48.0\text{ }\Omega

.

Step-by-Step Solution

1

Formulate the initial total resistance of the series circuit

R_{T1} = 8.0\text{ }\Omega + 40.0\text{ }\Omega + r = 48.0 + r

Before the shunt is added, the fixed resistor, galvanometer, and internal resistance of the cell are all connected in series.

2

Calculate the effective resistance of the shunted galvanometer

R_p = \frac{40.0 \times 10.0}{40.0 + 10.0} = 8.0\text{ }\Omega

Connecting the shunt in parallel across the galvanometer forms a parallel network.

3

Formulate the final total circuit resistance after shunting

R_{T2} = 8.0\text{ }\Omega + 8.0\text{ }\Omega + r = 16.0 + r

The new circuit consists of the fixed resistor, the parallel shunted galvanometer combination, and the internal resistance in series.

4

Set up the ratio equation for total circuit currents based on the given

50\%

current increase

\frac{E}{16.0 + r} = 1.5 \left(\frac{E}{48.0 + r}\right)

An increase of

50\%

means

I_2 = 1.5 I_1

. According to Ohm's law, total current is inversely proportional to total circuit resistance for a constant EMF

E

.

5

Solve the algebraic equation for internal resistance

r

r = 48.0\text{ }\Omega

Cross-multiplying yields

48.0 + r = 24.0 + 1.5r

, leading directly to

0.5r = 24.0

, so

r = 48.0\text{ }\Omega

.

Key Concept

Internal Resistance and Circuit Modification via Galvanometer Shunting

Question 126Question →

A conducting wire of length $1.5\,\text{m}$ and initial resistance $8.0\,\Omega$ is stretched uniformly until its length is doubled while its total volume remains unchanged. If a steady potential difference of $16\,\text{V}$ is applied across the ends of the stretched wire, what is the electric current flowing through it?

0.50\,\text{A}

1.00\,\text{A}

2.00\,\text{A}

8.00\,\text{A}

Show answer & explanation

Answer: $0.50\,\text{A}$

Answer

The electric current flowing through the stretched wire is

0.50\,\text{A}

.

When a cylindrical wire is stretched to double its length (

L_2 = 2L_1

) while maintaining constant volume, its cross-sectional area halves (

A_2 = A_1/2

). From the resistance formula

R = \rho L / A

, doubling length and halving area causes resistance to increase by a factor of four (

R_2 = 4 R_1 = 32.0\,\Omega

). By Ohm's law (

I = V/R

), applying

16\,\text{V}

across

32.0\,\Omega

yields a current of

0.50\,\text{A}

.

Step-by-Step Solution

1

Relate volume conservation to cross-sectional area.

Since volume

V = A_1 L_1 = A_2 L_2

remains constant when

L_2 = 2L_1

, the new cross-sectional area is

A_2 = A_1 / 2

.

Stretching a wire increases its length while proportionally reducing its area to preserve volume.

2

Calculate the new resistance of the stretched wire.

The new resistance is

R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1 / 2} = 4 R_1 = 4 \times 8.0\,\Omega = 32.0\,\Omega

.

Resistance is directly proportional to length and inversely proportional to cross-sectional area.

3

Apply Ohm's law to calculate current.

I = \frac{V}{R_2} = \frac{16\,\text{V}}{32.0\,\Omega} = 0.50\,\text{A}

.

Electric current is calculated by dividing potential difference by total resistance.

Key Concept

Effect of wire stretching on electrical resistance under volume conservation

Question 127Question →

An electric ceiling fan operates on a $220\,\text{V}$ mains supply and draws a steady current of $0.5\,\text{A}$ . What is the electrical power rating of the fan?

110\,\text{W}

440\,\text{W}

220.5\,\text{W}

110\,\text{J}

Show answer & explanation

Answer: $110\,\text{W}$

Answer

The electrical power rating of the fan is

110\,\text{W}

.

Electrical power

P

delivered to a circuit element is given by

P = VI

, where

V

is potential difference and

I

is current. Substituting

220\,\text{V}

and

0.5\,\text{A}

gives

110\,\text{W}

.

Step-by-Step Solution

1

Identify the given physical quantities

Voltage

V = 220\,\text{V}

and current

I = 0.5\,\text{A}

.

These are the basic electrical parameters required to calculate power.

2

Apply the electrical power formula

P = V \times I

Electrical power is defined as the product of potential difference across a device and current flowing through it.

3

Substitute values and compute the power

P = 220\,\text{V} \times 0.5\,\text{A} = 110\,\text{W}

Multiplying potential difference by current yields the rate of energy conversion in Watts.

Key Concept

Electrical Power (

P = VI

)

Estimated Time:45s

Question 128Question →

A cell of electromotive force $E = 6.0\text{ V}$ and internal resistance $r = 1.0\text{ }\Omega$ is connected in series with a resistor $R$ and a shunted galvanometer. The galvanometer has a resistance of $90\text{ }\Omega$ and produces a full-scale deflection for a current of $2.0\text{ mA}$ . If the shunt resistance connected across the galvanometer is $10\text{ }\Omega$ , calculate the value of the series resistor $R$ , in ohms $(\Omega)$ , required for the galvanometer to show full-scale deflection.

Show answer & explanation

Answer: 290

Answer

The required value of the series resistor

R

is

290\text{ }\Omega

.

At full-scale deflection, a current of

2.0\text{ mA}

passes through the galvanometer, resulting in a potential difference of

V_g = 2.0 \times 10^{-3} \times 90 = 0.18\text{ V}

. Since the shunt is connected in parallel with the galvanometer, the current through the shunt is

I_s = \frac{0.18}{10} = 0.018\text{ A} = 18\text{ mA}

. Thus, the total current supplied by the cell is

I = 2\text{ mA} + 18\text{ mA} = 20\text{ mA} = 0.02\text{ A}

. The total equivalent resistance of the shunted galvanometer is

R_p = \frac{90 \times 10}{90 + 10} = 9.0\text{ }\Omega

. Applying Ohm's law to the total loop including internal resistance

r

, we have

E = I(R + R_p + r) \implies 6.0 = 0.02(R + 9.0 + 1.0) \implies R + 10.0 = 300 \implies R = 290\text{ }\Omega

.

Step-by-Step Solution

1

Calculate voltage across the galvanometer at full-scale deflection

V_g = 0.18\text{ V}

The potential difference across parallel branches is equal, and for full-scale deflection

I_g = 2.0\text{ mA}

.

2

Calculate the current passing through the shunt resistor

I_s = 18.0\text{ mA} = 0.018\text{ A}

Using Ohm's law across the shunt resistor

S = 10\text{ }\Omega

with

V_s = V_g = 0.18\text{ V}

.

3

Calculate the total circuit current provided by the cell

I = 20.0\text{ mA} = 0.020\text{ A}

By Kirchhoff's current law, the main current splits between the galvanometer and shunt.

4

Find the equivalent resistance of the shunted galvanometer and total circuit resistance

R_p = 9.0\text{ }\Omega

and total circuit resistance

R_{\text{total}} = 300\text{ }\Omega

Parallel resistance formula gives

R_p = 9.0\text{ }\Omega

, and

R_{\text{total}} = \frac{E}{I} = \frac{6.0\text{ V}}{0.020\text{ A}} = 300\text{ }\Omega

.

5

Solve for the unknown external series resistance

R

R = 290\text{ }\Omega

R_{\text{total}} = R + r + R_p \implies 300 = R + 1.0 + 9.0 \implies R = 290\text{ }\Omega

.

Key Concept

Galvanometer Shunting and Electric Circuit Analysis with Internal Resistance

Question 129Question →

A tungsten filament lamp operating at a room temperature of $20\,^\circ\text{C}$ draws a current of $0.50\,\text{A}$ when connected to a $120\,\text{V}$ power source. When the lamp reaches its steady operating temperature, the current drops to $0.10\,\text{A}$ under the same voltage. If the temperature coefficient of resistance of tungsten is $4.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ , what is the operating temperature of the filament?

1020\,^\circ\text{C}

1000\,^\circ\text{C}

270\,^\circ\text{C}

200\,^\circ\text{C}

Show answer & explanation

Answer: $1020\,^\circ\text{C}$

Answer

The operating temperature of the tungsten filament is

1020\,^\circ\text{C}

.

Using Ohm's law (

R = V/I

), the initial resistance at

20\,^\circ\text{C}

is

240\,\Omega

and the operating resistance is

1200\,\Omega

. Substituting these into

R_2 = R_1[1 + \alpha(T_2 - T_1)]

yields

1200 = 240[1 + 4.0\times 10^{-3}(T_2 - 20)]

. Solving gives

T_2 - 20 = 1000\,^\circ\text{C}

, so the operating temperature is

1020\,^\circ\text{C}

.

Step-by-Step Solution

1

Calculate the resistance of the filament at room temperature (

20\,^\circ\text{C}

) and at the operating temperature using Ohm's Law (

R = V/I

).

Cold resistance

R_1 = \frac{120\,\text{V}}{0.50\,\text{A}} = 240\,\Omega

; Hot resistance

R_2 = \frac{120\,\text{V}}{0.10\,\text{A}} = 1200\,\Omega

.

Ohm's law relates voltage, current, and resistance for a conductor.

2

Apply the temperature dependence equation of electrical resistance:

R_2 = R_1[1 + \alpha(T_2 - T_1)]

.

\frac{1200}{240} = 1 + (4.0 \times 10^{-3})(T_2 - 20) \Rightarrow 5 = 1 + (4.0 \times 10^{-3})(T_2 - 20)

.

Resistance increases linearly with temperature according to the material's temperature coefficient.

3

Solve for the temperature difference

\Delta T = T_2 - 20

and find the final temperature

T_2

.

4.0 \times 10^{-3}(T_2 - 20) = 4 \Rightarrow T_2 - 20 = \frac{4}{4.0 \times 10^{-3}} = 1000\,^\circ\text{C} \Rightarrow T_2 = 1020\,^\circ\text{C}

.

Adding the initial temperature to the temperature change gives the final absolute operating temperature.

Key Concept

Temperature Dependence of Electrical Resistance and Ohm's Law

Question 130Question →

Two capacitors of capacitances $8.0\text{ }\mu\text{F}$ and $4.0\text{ }\mu\text{F}$ are connected in parallel. This parallel combination is then connected in series with a single $4.0\text{ }\mu\text{F}$ capacitor across a $36.0\text{ V}$ d.c. voltage source. What is the potential difference across the single $4.0\text{ }\mu\text{F}$ capacitor?

9.0\text{ V}

14.4\text{ V}

27.0\text{ V}

36.0\text{ V}

Show answer & explanation

Answer: $27.0\text{ V}$

Answer

The potential difference across the single

4.0\text{ }\mu\text{F}

capacitor is

27.0\text{ V}

.

The two parallel capacitors (

8.0\text{ }\mu\text{F}

and

4.0\text{ }\mu\text{F}

) combine directly to give an equivalent capacitance of

12.0\text{ }\mu\text{F}

. This equivalent capacitor is in series with the single

4.0\text{ }\mu\text{F}

capacitor across the

36.0\text{ V}

source. Using the voltage divider rule for series capacitors, the voltage across the single

4.0\text{ }\mu\text{F}

capacitor is

V = 36.0 \times \frac{12.0}{4.0 + 12.0} = 27.0\text{ V}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two parallel capacitors.

C_p = 8.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add algebraically.

2

Calculate the total equivalent capacitance of the entire circuit.

C_T = \frac{4.0 \times 12.0}{4.0 + 12.0} = \frac{48.0}{16.0} = 3.0\text{ }\mu\text{F}

The single

4.0\text{ }\mu\text{F}

capacitor and the

12.0\text{ }\mu\text{F}

parallel combination are in series.

3

Find the total charge supplied by the battery.

Q = C_T \times V = 3.0\text{ }\mu\text{F} \times 36.0\text{ V} = 108.0\text{ }\mu\text{C}

The total charge is equal to the total capacitance multiplied by the total voltage.

4

Calculate the potential difference across the single

4.0\text{ }\mu\text{F}

capacitor.

V_1 = \frac{Q}{C_1} = \frac{108.0\text{ }\mu\text{C}}{4.0\text{ }\mu\text{F}} = 27.0\text{ V}

In a series connection, the charge on the single capacitor equals the total charge.

Key Concept

Potential difference distribution in mixed series-parallel capacitor networks

Estimated Time:1m 30s

Question 131Question →

Match each physical scenario or effect involving magnetic forces listed on the left with its corresponding physical characteristic or outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair stationary charges with zero force; perpendicular entry with circular motion; anti-parallel currents with mutual repulsion; and force direction with mutual perpendicularity to velocity and magnetic field vectors.

Each physical scenario correctly pairs with its corresponding principle: stationary charges experience zero magnetic force; perpendicular charge motion forms a circular orbit; anti-parallel currents produce repulsion; and magnetic force is always mutually perpendicular to velocity and field vectors.

Step-by-Step Solution

1

Evaluate the magnetic force formula for a static charge

Using

F = qvB\sin\theta

, with

v = 0

,

F = 0

.

A magnetic field does not exert force on a stationary electric charge.

2

Determine the path of a particle moving perpendicular to a magnetic field

The magnetic force acts continuously at right angles to the velocity vector, providing centripetal acceleration and forming a circular orbit.

A perpendicular force of constant magnitude changes motion direction continuously without altering speed.

3

Apply Ampere's law and the right-hand rule to parallel currents in opposite directions

The magnetic field generated by each wire exerts an outward force on the other, producing repulsion.

Opposite currents create reinforcing field lines between the wires, driving them apart.

4

Analyze vector cross product orientation for magnetic force on a charge

The force vector

\vec{F}

is oriented perpendicular to the plane containing vectors

\vec{v}

and

\vec{B}

.

By definition of vector cross-product

\vec{F} = q(\vec{v} \times \vec{B})

, the resultant vector is orthogonal to both input vectors.

Key Concept

Magnetic Force on Moving Charges and Current-Carrying Conductors

Question 132Question →

An RLC series circuit connected across a $100\text{ V}$ (RMS) AC voltage source operates at resonance, dissipating an average power of $400\text{ W}$ . If the inductive reactance at resonance is $25\ \Omega$ , what is the total impedance of the circuit when the capacitance is adjusted such that the capacitive reactance increases by $60\ \Omega$ ?

65\ \Omega

85\ \Omega

60\ \Omega

110\ \Omega

Show answer & explanation

Answer: $65\ \Omega$

Answer

The total impedance of the circuit after adjusting the capacitance is

65\ \Omega

.

At resonance, the net reactance is zero and the circuit behaves purely resistively. Using

P = \frac{V_{\text{rms}}^2}{R}

, the resistance is

R = \frac{100^2}{400} = 25\ \Omega

. Since

X_L = X_C = 25\ \Omega

at resonance, increasing capacitive reactance by

60\ \Omega

gives a new capacitive reactance of

85\ \Omega

. The net reactance magnitude is

|25\ \Omega - 85\ \Omega| = 60\ \Omega

. Combining resistance and net reactance in quadrature yields an impedance of

Z = \sqrt{25^2 + 60^2} = 65\ \Omega

.

Step-by-Step Solution

1

Determine the resistance of the circuit at resonance using the power dissipation formula.

At resonance, impedance equals resistance (

Z = R

) and phase angle is zero, so

P = \frac{V_{\text{rms}}^2}{R} \implies R = \frac{(100)^2}{400} = 25\ \Omega

.

At resonance, inductive reactance and capacitive reactance cancel each other out completely.

2

Identify the initial and modified reactances.

At resonance,

X_L = X_C = 25\ \Omega

. After adjustment, the new capacitive reactance is

X_C' = 25\ \Omega + 60\ \Omega = 85\ \Omega

.

Capacitive reactance was increased by

60\ \Omega

from its resonant value.

3

Calculate the net reactance of the modified circuit.

X_{\text{net}} = |X_L - X_C'| = |25\ \Omega - 85\ \Omega| = 60\ \Omega

.

Net reactance is the magnitude of the difference between inductive and capacitive reactances.

4

Calculate the new total impedance using phasor addition.

Z = \sqrt{R^2 + (X_L - X_C')^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65\ \Omega

.

Resistance and net reactance are

90^\circ

out of phase, requiring vector summation (Pythagorean theorem).

Key Concept

Resonance and Impedance in AC Circuits

Question 133Question →

Two concentric circular conducting loops, $P$ and $Q$ , lie flat in the same horizontal plane, with loop $Q$ situated inside loop $P$ . Loop $P$ is connected to a DC power source through a variable resistor and initially carries a steady clockwise current. If the resistance of the variable resistor is suddenly decreased, which of the following correctly describes the direction of the induced current in loop $Q$ and the nature of the magnetic force exerted on loop $Q$ ?

The induced current in

Q

is counter-clockwise, and the magnetic force on

Q

is repulsive.

The induced current in

Q

is clockwise, and the magnetic force on

Q

is attractive.

The induced current in

Q

is counter-clockwise, and the magnetic force on

Q

is attractive.

The induced current in

Q

is clockwise, and the magnetic force on

Q

is repulsive.

Show answer & explanation

Answer: The induced current in $Q$ is counter-clockwise, and the magnetic force on $Q$ is repulsive.

Answer

The induced current in loop

Q

flows in a counter-clockwise direction, and the magnetic force between loop

P

and loop

Q

is repulsive.

Decreasing the variable resistance increases the clockwise current in loop

P

, which increases the downward magnetic flux through loop

Q

. By Lenz's law, loop

Q

generates an opposing upward magnetic flux, which corresponds to a counter-clockwise induced current. Because the two loops carry currents flowing in opposite directions, they exert a repulsive magnetic force on each other.

Step-by-Step Solution

1

Determine the initial magnetic field direction created by loop

P

Using the right-hand grip rule, a clockwise current in outer loop

P

produces a magnetic field directed perpendicularly downward (into the plane of the page) inside the loop.

Current in a circular loop generates an axial magnetic field whose direction is given by the right-hand rule.

2

Analyze the change in magnetic flux through inner loop

Q

Decreasing the resistance increases the current in loop

P

, thereby increasing the downward magnetic flux passing through loop

Q

.

Ohm's law (

I = V/R

) indicates that reducing resistance increases current, which proportionally strengthens the magnetic field (

B \propto I

).

3

Apply Faraday's and Lenz's laws to find the induced current direction in

Q

To oppose the increasing downward flux, the induced current in loop

Q

must produce a magnetic field directed upward (out of the page). By the right-hand grip rule, an upward field requires a counter-clockwise current in loop

Q

.

Lenz's law states that an induced current always flows in such a direction that its magnetic effect opposes the change producing it.

4

Determine the nature of the magnetic force between the two loops

Loop

P

carries a clockwise current while loop

Q

carries a counter-clockwise current. Concentric circular conductors carrying currents in opposite directions repel each other.

Opposing electric currents produce magnetic fields that result in mutual magnetic repulsion.

Key Concept

Lenz's Law and Electromagnetic Induction in Coaxial Loops

Question 134Question →

An alternating current (AC) circuit contains an inductor of inductance $L = \frac{0.2}{\pi^2}\ \text{H}$ , a resistor of resistance $R = 40\ \Omega$ , and a variable capacitor $C$ connected in series across a $50\ \text{Hz}$ voltage supply. What capacitance $C$ , in microfarads ( $\mu\text{F}$ ), is required for the circuit to operate at electrical resonance?

Show answer & explanation

Answer: 500

Answer

The capacitance required to achieve electrical resonance is 500 μF.

At electrical resonance in a series RLC circuit, the inductive reactance (

X_L

) equals the capacitive reactance (

X_C

). Setting

2\pi f L = \frac{1}{2\pi f C}

yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

f_0 = 50\ \text{Hz}

and

L = \frac{0.2}{\pi^2}\ \text{H}

into this equation gives

C = 5 \times 10^{-4}\ \text{F}

, which equals

500\ \mu\text{F}

.

Step-by-Step Solution

1

Recall the resonant frequency formula for a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

).

2

Substitute the given numerical parameters into the equation.

50 = \frac{1}{2\pi \sqrt{\frac{0.2}{\pi^2} \cdot C}}

Given frequency

f_0 = 50\ \text{Hz}

and inductance

L = \frac{0.2}{\pi^2}\ \text{H}

.

3

Isolate the square root term and simplify.

\sqrt{0.2 C} = 0.01

Simplifying

2\pi \cdot \frac{1}{\pi} = 2

and rearranging

2 \sqrt{0.2 C} = \frac{1}{50} = 0.02

.

4

Square both sides and solve for

C

in farads.

C = 5 \times 10^{-4}\ \text{F}

0.2 C = (0.01)^2 = 10^{-4}

, so

C = \frac{10^{-4}}{0.2} = 5 \times 10^{-4}\ \text{F}

.

5

Convert capacitance from farads to microfarads.

C = 500\ \mu\text{F}

Multiply farads by

10^6

to express the result in microfarads.

Key Concept

Resonant Frequency in Series AC Circuits

Question 135Question →

An RLC series circuit operating at resonance contains an inductor of inductance $0.1\text{ H}$ and a capacitor of capacitance $10\ \mu\text{F}$ . What is the resonant angular frequency of the circuit in radians per second ( $\text{rad/s}$ )?

Show answer & explanation

Answer: 1000

Answer

The resonant angular frequency of the circuit is

1000\text{ rad/s}

.

The resonant angular frequency

\omega_0

of an AC circuit is determined by the formula

\omega_0 = \frac{1}{\sqrt{LC}}

. Substituting the given values

L = 0.1\text{ H}

and

C = 10^{-5}\text{ F}

gives

\omega_0 = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

Step-by-Step Solution

1

Convert given parameters to standard SI units

L = 0.1\text{ H}

and

C = 10 \times 10^{-6}\text{ F} = 10^{-5}\text{ F}

.

Calculations must be performed in base SI units for dimensional consistency.

2

Calculate the resonant angular frequency using

\omega_0 = \frac{1}{\sqrt{LC}}

\omega_0 = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

), which yields

\omega_0 = \frac{1}{\sqrt{LC}}

.

Key Concept

Resonant Angular Frequency

Question 136Question →

An alternating voltage source of root-mean-square (RMS) voltage $200\ \text{V}$ is connected across a series combination of a resistor with resistance $60\ \Omega$ and a capacitor with capacitive reactance $80\ \Omega$ . Calculate the root-mean-square current in the circuit.

Show answer & explanation

Answer: 2

Answer

The root-mean-square current flowing through the circuit is 2 A.

The total impedance of the series RC circuit is obtained via quadrature sum

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = 100\ \Omega

. Dividing the RMS supply voltage (

200\ \text{V}

) by this impedance gives an RMS current of

2\ \text{A}

.

Step-by-Step Solution

1

Calculate the total impedance of the series RC circuit.

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = 100\ \Omega

In an AC circuit with resistance and capacitive reactance in series, the total Opposition (impedance) is found by phasor addition.

2

Apply Ohm's law for alternating current circuits to find RMS current.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{100\ \Omega} = 2\ \text{A}

The RMS current is equal to the RMS voltage divided by the total circuit impedance.

Key Concept

Impedance and RMS Current in AC Series Circuits

Question 137Question →

A point charge $Q = +6.0 \times 10^{-9}\text{ C}$ is fixed in a vacuum. A small test charge $q = +2.0 \times 10^{-9}\text{ C}$ is placed at a distance of $0.30\text{ m}$ from $Q$ . What is the magnitude of the electric field intensity produced by $Q$ at the location of the test charge? [Take $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ ]

1.8 \times 10^2\text{ N C}^{-1}

6.0 \times 10^2\text{ N C}^{-1}

8.0 \times 10^2\text{ N C}^{-1}

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

Show answer & explanation

Answer: $6.0 \times 10^2\text{ N C}^{-1}$

Answer

6.0 \times 10^2\text{ N C}^{-1}

The magnitude of the electric field intensity

E

created by a point charge

Q

at a distance

r

is given by

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

. Substituting

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

,

Q = 6.0 \times 10^{-9}\text{ C}

, and

r = 0.30\text{ m}

yields

E = \frac{54}{0.09} = 600\text{ N C}^{-1}

, which equals

6.0 \times 10^2\text{ N C}^{-1}

. The test charge magnitude does not affect the electric field produced by the source charge.

Step-by-Step Solution

1

Identify the formula for electric field intensity due to a point charge

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

The electric field intensity depends solely on the magnitude of the source charge

Q

and the square of the distance

r

from it.

2

Substitute the given values into the formula

E = \frac{(9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (6.0 \times 10^{-9}\text{ C})}{(0.30\text{ m})^2}

Insert

k = 9.0 \times 10^9

, source charge

Q = 6.0 \times 10^{-9}\text{ C}

, and distance

r = 0.30\text{ m}

.

3

Simplify the arithmetic

E = \frac{54}{0.09} = 600\text{ N C}^{-1} = 6.0 \times 10^2\text{ N C}^{-1}

Squaring

0.30

gives

0.09

, and dividing

54

by

0.09

yields

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

.

Key Concept

Electric Field Intensity of a Point Charge

Question 138Question →

A straight copper rod of length $0.50\text{ m}$ and mass $40\text{ g}$ carries a steady current and is suspended horizontally in a uniform magnetic field of $0.40\text{ T}$ . The field is directed horizontally at an angle of $30^\circ$ to the length of the rod. If the upward magnetic force acting on the rod exactly balances its weight, what is the magnitude of the current flowing through the rod? (Take $g = 10\text{ m/s}^2$ )

0.25\text{ A}

2.0\text{ A}

4.0\text{ A}

40\text{ A}

Show answer & explanation

Answer: $4.0\text{ A}$

Answer

The magnitude of the current required to balance the weight of the rod is

4.0\text{ A}

.

For the rod to remain suspended in equilibrium, the upward magnetic force

F_B = BIL \sin \theta

must balance the downward gravitational force

W = mg

. Substituting

m = 0.040\text{ kg}

,

g = 10\text{ m/s}^2

,

B = 0.40\text{ T}

,

L = 0.50\text{ m}

, and

\theta = 30^\circ

yields

0.40 = 0.40 \times I \times 0.50 \times 0.50

, which solves to

I = 4.0\text{ A}

.

Step-by-Step Solution

1

Convert the mass of the rod to kilograms and calculate its weight.

m = 40\text{ g} = 0.040\text{ kg}

, so

W = mg = 0.040\text{ kg} \times 10\text{ m/s}^2 = 0.40\text{ N}

.

Standard SI units must be used to ensure dimensional consistency.

2

Express the magnetic force acting on the conductor in terms of current

I

.

F_B = B I L \sin \theta = (0.40\text{ T}) \times I \times (0.50\text{ m}) \times \sin 30^\circ = 0.10 I\text{ N}

.

The magnetic force on a current-carrying conductor at an angle

\theta

to a magnetic field is given by

F = BIL \sin \theta

.

3

Equate the upward magnetic force to the downward weight to find

I

.

0.10 I = 0.40 \implies I = \frac{0.40}{0.10} = 4.0\text{ A}

.

For vertical equilibrium, the net vertical force must equal zero.

Key Concept

Equilibrium of a current-carrying conductor in a uniform magnetic field (

F = BIL \sin \theta = mg

)

Estimated Time:2m 0s

Question 139Question →

At the Earth's magnetic equator, a freely suspended magnetic dip needle comes to rest horizontally, resulting in an angle of dip of $0^\circ$ .

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the Earth's magnetic field lines at the magnetic equator are parallel to the Earth's surface. Consequently, the vertical component of the magnetic field is zero, causing a freely suspended dip needle to settle completely horizontally at an angle of

0^\circ

.

Step-by-Step Solution

1

Define the angle of dip (magnetic inclination).

The angle of dip

\theta

is the angle between the Earth's magnetic field direction and the horizontal plane at a given location.

This definition helps determine how a dip needle orientates itself relative to the surface of the Earth.

2

Examine the Earth's magnetic field components at the magnetic equator.

At the magnetic equator, the vertical component of Earth's magnetic field is zero (

B_v = 0

), so the total field is entirely horizontal (

B = B_h

).

The magnetic flux lines are parallel to the geographical surface along the magnetic equator.

3

Evaluate the statement.

Since the field vector is horizontal, the angle of dip

\theta = \tan^{-1}\left(\frac{B_v}{B_h}\right) = \tan^{-1}(0) = 0^\circ

, making the statement True.

The dip needle aligns with the total magnetic field vector, resting horizontally at

0^\circ

.

Key Concept

Angle of Dip at the Magnetic Equator

Question 140Question →

Match each physical phenomenon or quantity involving electromagnetic forces listed on the left with its corresponding governing mathematical equation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Magnetic force on a current-carrying conductor pairs with

F = B I L \sin \theta

; Force per unit length between parallel conductors pairs with

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

; Radius of circular trajectory of a charged particle pairs with

r = \frac{m v}{q B}

; Torque on a current-carrying coil pairs with

\tau = B I A N \sin \theta

.

Each electromagnetic phenomenon matches directly with its derived expression from the magnetic force laws: magnetic force on a wire is

F = B I L \sin \theta

, force between parallel wires is

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

, orbit radius of charge is

r = \frac{m v}{q B}

, and coil torque is

\tau = B I A N \sin \theta

.

Step-by-Step Solution

1

Analyze the magnetic force on a straight conductor

The Lorentz force law applied to current elements yields

F = B I L \sin \theta

.

Free charges moving inside the wire experience magnetic force perpendicular to both current and magnetic field vector.

2

Determine the mutual force formula for parallel conductors

The magnetic field from wire 1 is

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

, giving force per length

\frac{F}{L} = B_1 I_2 = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Each conductor sits within the circular magnetic field lines generated by the other conductor.

3

Derive the motion equation for a charged particle in a magnetic field

Setting

q v B = \frac{m v^2}{r}

yields

r = \frac{m v}{q B}

.

The magnetic force provides the required inward centripetal acceleration for circular motion.

4

Identify the expression for torque on a magnetic dipole / coil

The couple produced by forces on opposite sides of a rectangular loop gives

\tau = B I A N \sin \theta

.

Opposite sides experience forces in opposing directions separated by a moment arm.

Key Concept

Formulas for magnetic forces on current-carrying conductors, moving charges, parallel wires, and coils

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