Electromagnetic Induction Practice Questions — JAMB UTME

Question 1Question →

A flat circular coil consisting of $100$ turns and having a cross-sectional area of $4.0 \times 10^{-3}\text{ m}^2$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.02\text{ s}$ , what is the magnitude of the average electromotive force (e.m.f.) induced in the coil?

10\text{ V}

20\text{ V}

0.20\text{ V}

40\text{ V}

Show answer & explanation

Answer: $20\text{ V}$

Answer

20\text{ V}

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force equals the rate of change of magnetic flux linkage. Because the magnetic field direction is completely reversed, the change in field magnitude is

\Delta B = 0.50 - (-0.50) = 1.0\text{ T}

. Multiplying by the area (

4.0 \times 10^{-3}\text{ m}^2

) and turn count (

100

), then dividing by the duration (

0.02\text{ s}

) gives

20\text{ V}

.

Step-by-Step Solution

1

Determine the change in magnetic flux density (

\Delta B

)

\Delta B = B_f - B_i = -0.50\text{ T} - 0.50\text{ T} = -1.0\text{ T}

, so

|\Delta B| = 1.0\text{ T}

Reversing the magnetic field direction changes the sign of the flux density relative to the coil area.

2

Calculate the total change in magnetic flux linkage (

\Delta \Phi_{total}

)

\Delta \Phi_{total} = N \cdot A \cdot |\Delta B| = 100 \times (4.0 \times 10^{-3}\text{ m}^2) \times 1.0\text{ T} = 0.40\text{ Wb-turns}

The total magnetic flux linkage is proportional to the number of turns, the surface area, and the net field change.

3

Apply Faraday's law of electromagnetic induction to find the induced e.m.f. (

\mathcal{E}

)

\mathcal{E} = \frac{\Delta \Phi_{total}}{\Delta t} = \frac{0.40\text{ Wb-turns}}{0.02\text{ s}} = 20\text{ V}

The induced e.m.f. magnitude is equal to the rate of change of magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 2Question →

A rectangular coil of $100$ turns with dimensions $0.10\text{ m}$ by $0.20\text{ m}$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . The coil is rotated through $90^\circ$ about an axis perpendicular to the field lines in a time interval of $0.040\text{ s}$ , bringing its plane parallel to the magnetic field. What is the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts?

Show answer & explanation

Answer: 25

Answer

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

25\text{ V}

.

According to Faraday's law, the induced electromotive force magnitude is

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

. The initial flux through each turn is

\Phi_1 = B A = 0.50\text{ T} \times 0.020\text{ m}^2 = 0.010\text{ Wb}

. When rotated parallel to the field, the final flux

\Phi_2

is

0\text{ Wb}

, so

\Delta \Phi = 0.010\text{ Wb}

. Substituting

N = 100

and

\Delta t = 0.040\text{ s}

yields an induced e.m.f. of

100 \times \frac{0.010}{0.040} = 25\text{ V}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the rectangular coil.

A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2

The surface area is required to find the magnetic flux passing through each turn.

2

Determine the change in magnetic flux through one turn of the coil.

Initial flux

\Phi_1 = B A = 0.50 \times 0.020 = 0.010\text{ Wb}

; Final flux

\Phi_2 = 0\text{ Wb}

; Change

\Delta \Phi = 0.010\text{ Wb}

When perpendicular to the magnetic field, maximum flux links the coil. Rotating it parallel reduces the flux linking the coil to zero.

3

Apply Faraday's law of electromagnetic induction to calculate induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 100 \times \frac{0.010\text{ Wb}}{0.040\text{ s}} = 25\text{ V}

Faraday's law states that the induced e.m.f. magnitude equals the rate of change of total magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 3Question →

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 4Question →

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 5Question →

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 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

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 11Question →

According to Lenz's law, what occurs at the near face of a stationary solenoid when the north pole of a bar magnet is moved rapidly towards it?

A magnetic north pole is induced at the near face to oppose the approaching magnet.

A magnetic south pole is induced at the near face to attract the approaching magnet.

A magnetic south pole is induced at the near face to repel the approaching magnet.

No magnetic polarity is produced because induction generates only electric voltage without magnetic fields.

Show answer & explanation

Answer: A magnetic north pole is induced at the near face to oppose the approaching magnet.

Answer

A magnetic north pole is induced at the near face to oppose the approaching magnet.

Lenz's law states that the direction of an induced current is always such that its magnetic effect opposes the motion or change causing it. When a north pole approaches the solenoid face, the induced current flows counter-clockwise (viewed from the magnet) to form an induced magnetic north pole at that face, thereby exerting a repulsive force that opposes the motion.

Step-by-Step Solution

1

Identify the cause of change in magnetic flux.

The approaching north pole increases magnetic flux linking the solenoid.

Electromagnetic induction occurs whenever magnetic flux linked with a conductor changes.

2

Apply Lenz's law to determine the polarity of the induced field.

The induced current must produce a magnetic field that opposes the increase in flux caused by the approaching north pole.

Lenz's law states that the induced effect always opposes the cause producing it.

3

Deduce the required magnetic pole at the near face.

A north pole must be set up at the near face because like magnetic poles repel each other, opposing the inward motion.

Repulsion between like poles (North against North) exerts a retarding force on the approaching magnet.

Key Concept

Lenz's Law of Electromagnetic Induction

Question 12Question →

A step-down transformer connected to a $240\text{ V}$ AC mains supply operates a $12\text{ V}, 48\text{ W}$ lamp at its normal brightness rating. If the efficiency of the transformer is $80\%$ , what is the electric current drawn by the primary winding from the mains supply?

0.16\text{ A}

0.20\text{ A}

0.25\text{ A}

4.00\text{ A}

Show answer & explanation

Answer: $0.25\text{ A}$

Answer

0.25\text{ A}

The output power delivered to the lamp is

48\text{ W}

. Accounting for the transformer's

80\%

efficiency, the input power required at the primary winding is

\frac{48\text{ W}}{0.80} = 60\text{ W}

. Since the primary voltage is

240\text{ V}

, the primary current drawn is

I_p = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

.

Step-by-Step Solution

1

Determine the power output at the secondary winding

P_s = 48\text{ W}

The lamp operates at its normal rating, so secondary power equals the lamp rating.

2

Calculate the required power input to the primary winding using transformer efficiency

P_p = \frac{P_s}{\eta} = \frac{48\text{ W}}{0.80} = 60\text{ W}

Efficiency is defined as

\eta = \frac{P_s}{P_p}

, meaning primary input power must exceed secondary output power due to losses.

3

Compute the primary current drawn from the supply

I_p = \frac{P_p}{V_p} = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

Power in an AC primary circuit is given by

P_p = V_p I_p

assuming a purely resistive secondary load.

Key Concept

Transformer Efficiency and Power Transfer in Electromagnetic Induction

Question 13Question →

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 14Question →

A bar magnet is pulled away from a stationary circular coil such that its South pole moves directly away from the front face of the coil. Based on Lenz's law, which polarity is induced on the front face of the coil?

North pole, to attract the receding South pole and oppose its motion

South pole, to repel the receding South pole and oppose its motion

South pole, to attract the receding South pole and assist its motion

North pole, to repel the receding South pole and assist its motion

Show answer & explanation

Answer: North pole, to attract the receding South pole and oppose its motion

Answer

A North pole is induced on the front face of the coil to attract the receding South pole and oppose its motion.

According to Lenz's law, the direction of induced current creates a magnetic field that opposes the change causing it. As the South pole moves away, the decreasing magnetic flux is opposed by an attractive force pulling the magnet back. Therefore, an opposite pole (North pole) is induced on the coil's front face.

Step-by-Step Solution

1

Identify the change causing electromagnetic induction

The South pole of the magnet is moving away from the coil, causing a decrease in magnetic flux through the coil.

Induction is driven by a change in magnetic flux according to Faraday's law.

2

Apply Lenz's law to determine the direction of induced magnetic effect

The induced magnetic field must attempt to pull the magnet back to oppose its withdrawal.

Lenz's law dictates that the direction of an induced current always opposes the motion or change causing it.

3

Determine the required magnetic polarity on the front face

To attract the departing South pole, an opposite magnetic pole (North pole) must be induced on the front face.

Unlike magnetic poles attract each other.

Key Concept

Lenz's Law and Direction of Induced Current

Estimated Time:45s

Question 15Question →

A flat circular coil of $80$ turns, each having an area of $0.02\text{ m}^2$ , is placed perpendicularly in a uniform magnetic field. If the magnetic flux density decreases uniformly from $0.60\text{ T}$ to $0\text{ T}$ in $0.16\text{ s}$ , what is the magnitude of the induced electromotive force (e.m.f.) in the coil?

0.075\text{ V}

6.0\text{ V}

9.6\text{ V}

15.0\text{ V}

Show answer & explanation

Answer: $6.0\text{ V}$

Answer

6.0\text{ V}

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is equal to the rate of change of magnetic flux linkage:

\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N \cdot A \cdot \frac{\Delta B}{\Delta t}

. Substituting

N = 80

,

A = 0.02\text{ m}^2

,

\Delta B = 0.60\text{ T}

, and

\Delta t = 0.16\text{ s}

yields

\mathcal{E} = 80 \times 0.02 \times \frac{0.60}{0.16} = 6.0\text{ V}

.

Step-by-Step Solution

1

Calculate the change in magnetic flux density (

\Delta B

) and the change in magnetic flux per turn (

\Delta \Phi

).

\Delta B = 0.60\text{ T} - 0\text{ T} = 0.60\text{ T}

, and

\Delta \Phi = A \times \Delta B = 0.02\text{ m}^2 \times 0.60\text{ T} = 0.012\text{ Wb}

.

Magnetic flux is defined as the product of the perpendicular magnetic flux density and the cross-sectional area.

2

Apply Faraday's Law of Electromagnetic Induction for an

N

-turn coil:

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

.

\mathcal{E} = 80 \times \frac{0.012\text{ Wb}}{0.16\text{ s}} = 80 \times 0.075\text{ V} = 6.0\text{ V}.

The induced e.m.f. is directly proportional to the total rate of change of magnetic flux linkage through all

N

turns of the coil.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 16Question →

A long straight conductor lying in the plane of the page carries a steady current directed towards the top of the page. A rectangular conducting loop lies in the same plane to the right of the conductor. If the loop is pulled horizontally to the right away from the conductor, what is the direction of the induced current in the loop and the direction of the net magnetic force exerted on the loop?

Clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Counter-clockwise induced current and a net magnetic force directed to the right (away from the conductor)

Clockwise induced current and a net magnetic force directed to the right (away from the conductor)

Counter-clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Show answer & explanation

Answer: Clockwise induced current and a net magnetic force directed to the left (towards the conductor)

Answer

The induced current flows in a clockwise direction, and the net magnetic force acts to the left (towards the straight conductor).

The straight wire creates a magnetic field pointing into the page on its right side. Pulling the loop further away decreases the inward magnetic flux passing through it. By Lenz's law, the induced current must create an inward magnetic field to oppose this decrease, which corresponds to a clockwise current. Furthermore, Lenz's law requires the resulting mechanical magnetic force to oppose the rightward motion, producing a net force directed to the left (towards the conductor).

Step-by-Step Solution

1

Determine magnetic field direction around the long straight wire

Using the right-hand grip rule, the magnetic field

\vec{B}

produced by the upward current to the right of the wire points perpendicularly into the page.

Current flowing upward creates concentric magnetic field lines that enter the plane of the page on the right side.

2

Analyze magnetic flux change when moving the loop away

As the loop moves rightward away from the wire, it enters a region of weaker magnetic field, so magnetic flux pointing into the page decreases.

Magnetic field strength decreases inversely with distance (

B \propto 1/r

).

3

Apply Lenz's Law to find induced current direction

The induced current must oppose the decrease in magnetic flux into the page by producing its own magnetic field directed into the page. By the right-hand rule for loops, a clockwise current produces an inward field.

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

4

Determine the net magnetic force on the loop

According to Lenz's law, the net force must oppose the motion causing the induction. Since the loop moves right, the net force must point left (towards the conductor).

Alternatively, the left edge of the loop is closer to the wire and carries upward current (clockwise loop), parallel to the main wire's upward current. Parallel currents attract, yielding a net force to the left.

Key Concept

Lenz's Law and Electromagnetic Induction

Alternative Method

Consider magnetic forces on parallel current segments: The left side of the rectangular loop has current flowing upward (for clockwise flow), which is parallel to the main wire's upward current and is therefore attracted to the left. The right side has downward current (antiparallel) and is repelled to the right. Because the left side is closer to the wire, the attractive force dominates, yielding a net force to the left.

Estimated Time:1m 15s

Question 17Question →

When a soft-iron core is inserted into an air-core solenoid connected to a constant-voltage alternating current (AC) source, the root-mean-square (rms) current flowing through the circuit increases.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Inserting a soft-iron core increases the self-inductance and inductive reactance of the coil, which decreases the rms current.

Inserting a soft-iron core raises the magnetic permeability, thereby increasing the coil's self-inductance (

L

) and inductive reactance (

X_L = 2\pi f L

). This higher reactance opposes current flow more strongly, causing the rms current to drop.

Step-by-Step Solution

1

Determine the effect of a soft-iron core on self-inductance.

Soft iron has high magnetic permeability, which concentrates magnetic flux lines and markedly increases the coil's self-inductance (

L

).

Self-inductance is directly proportional to the magnetic permeability of the core material.

2

Relate self-inductance to inductive reactance in an AC circuit.

Inductive reactance is given by

X_L = 2\pi f L

. An increase in

L

produces a proportional increase in

X_L

.

Inductive reactance represents the opposition offered by an inductor to alternating current.

3

Calculate the impact on root-mean-square (rms) current.

Using Ohm's law for AC reactive circuits,

I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L}

. Since

X_L

increases,

I_{\text{rms}}

decreases.

Current is inversely proportional to reactance for a constant supply voltage.

Key Concept

Self-Inductance and Inductive Reactance in AC Circuits

Estimated Time:1m 0s

Question 18Question →

When a straight metallic conductor of length $L$ moves at a constant velocity $v$ perpendicular to a uniform magnetic field $B$ , free electrons inside the conductor accumulate at one end, creating an internal electric field that eventually balances the magnetic force acting on them.

True

False

Show answer & explanation

Answer: True

Answer

True. Moving a conductor through a magnetic field exerts a Lorentz magnetic force on its free electrons, pushing them toward one end. This separation of charge produces an internal electric field that exerts an opposing electrostatic force, reaching equilibrium when

qE = qvB

and resulting in an induced motional e.m.f. of

\mathcal{E} = BLv

.

The statement is correct because free charge carriers in a conductor moving through a magnetic field experience a magnetic Lorentz force. This force drives electrons to one side of the conductor, leaving positive ions on the other. The resulting charge separation builds an internal electric field

E

until the electric force

qE

equals the magnetic force

qvB

, creating a stable motional e.m.f.

\mathcal{E} = BLv

.

Step-by-Step Solution

1

Analyze the force acting on free electrons due to motion in a magnetic field.

Each free electron carrying charge

q

experiences a magnetic force of magnitude

F_m = qvB

directed along the length of the conductor.

According to the Lorentz force law, a charge moving with velocity

v

perpendicular to a magnetic field

B

experiences a magnetic force perpendicular to both motion and magnetic field.

2

Determine the consequence of electron movement within the conductor.

Electrons accumulate at one end, making that end negatively charged and leaving the opposite end positively charged.

The conductor has finite boundaries, so mobile charge carriers migrate until stopped by the physical ends of the rod.

3

Evaluate the electric field and equilibrium condition established by charge separation.

An electric field

E

is formed pointing from the positive end to the negative end, creating an opposing electrostatic force

F_e = qE

. Accumulation stops when

F_e = F_m

, leading to

E = vB

and motional e.m.f.

\mathcal{E} = EL = BLv

.

Steady-state motional e.m.f. requires electrostatic equilibrium between the magnetic force driving charges apart and the electric force pulling them back.

Key Concept

Motional Electromotive Force and Microscopic Charge Separation

Electromagnetic Induction

Answer

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