Electromagnetic Induction | Physics JEE Main questions with Solutions

Q 11 :

The induced emf can be produced in a coil by [2023]

A. moving the coil with uniform speed inside uniform magnetic field
B. moving the coil with non-uniform speed inside uniform magnetic field
C. rotating the coil inside the uniform magnetic field
D. changing the area of the coil inside the uniform magnetic field

Choose the correct answer from the options given below:

B and C only
B and D only
C and D only
A and C only

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(3)

Moving a coil inside a uniform magnetic field either with uniform or non-uniform speed doesn’t change flux, so no emf is induced.

Q 12 :

An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when it moves with a velocity of [2023]

$20 {ms}^{- 1}$
$2 {ms}^{- 1}$
$0.5 {ms}^{- 1}$
$3.2 {ms}^{- 1}$

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(2)

$Induced emf = \frac{B}{v}$

$\Rightarrow 0.08 = 0.4 (\frac{10}{100}) v$

$\Rightarrow v = (\frac{0.08 \times 10}{0.4}) \Rightarrow v = 2 m/s$

Q 13 :

A 1 m long metal rod XY completes the circuit as shown in the figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is $5 Ω$ , the force needed to move the rod in the direction, as indicated, with a constant speed of 4 m/s will be ________ $10^{- 3} N$ . [2023]

0%

(18)

$F = i l B$

$= (\frac{ε}{R}) l B = (\frac{v B l}{R}) l B = \frac{v B^{2} l^{2}}{R} = \frac{4}{5} \times {(\frac{15}{100})}^{2} \times 1^{2}$

$= \frac{4}{5} \times \frac{225}{10^{4}} = \frac{180}{10^{4}} = 0.018 N = 18 \times 10^{- 3} N$

Q 14 :

A metallic cube of side 15 cm moving along y-axis at a uniform velocity of $2 {ms}^{- 1}$ . In a region of uniform magnetic field of magnitude 0.5 T directed along z-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV. [2023]

0%

(150)

$Δ V = (v \times B) d$

$Δ V = (2 \times \frac{1}{2}) \times 0.15$

$Δ V = 150 mV$

Q 15 :

A 20 cm long metallic rod is rotated with 210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.2 T parallel to the axis exists everywhere. The emf developed between the centre and the ring is _________ mV. Take $π = \frac{22}{7}$ . [2023]

0%

(88)

$Here ω = 210 rpm = 210 \times \frac{2 π}{60} rad.s$

$\Rightarrow ω = 7 π rad/s. and l = 0.2 m$

$and B = 0.2 T$

$emf developed across rod is = \frac{1}{2} B ω l^{2}$

$\frac{1}{2} \times 0.2 \times 7 π \times {(0.2)}^{2} = 88 mV$

Q 16 :

In an AC generator, a rectangular coil of 100 turns each having area $14 \times 10^{- 2} m^{2}$ is rotated at 360 rev./min. about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T. The maximum value of the emf produced will be _________ V. (Take $π = \frac{22}{7}$ ) [2023]

0%

(1584)

$ξ_{\max} = N A B ω$

$= 100 \times 14 \times 10^{- 2} \times 3 \times \frac{360 \times 2 π}{60} = 1584 V$

Q 17 :

A 20 m long uniform copper wire held horizontally is allowed to fall under gravity $(g = 10 {m/s}^{2})$ through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travels a vertical distance of 200 m is ________ mV. [2026]

$20 \sqrt{10}$
$0.2 \sqrt{10}$
$200 \sqrt{10}$
$2 \sqrt{10}$

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(1)

$ε = v B ℓ$

$v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 200} = 20 \sqrt{10}$

$ε = (20 \sqrt{10}) (0.5 \times 10^{- 4}) 20$

$= 20 \sqrt{10} \times 10^{- 3} = 20 \sqrt{10} mV$

Q 18 :

A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60 °$ with the vertical, then the maximum induced EMF between the point of suspension and the point of oscillation is ______ mV. (Take $g = 10 {m/s}^{2}$ ). [2026]

0%

(100)

$ε_{\max} = \frac{B ω_{\max} l^{2}}{2} . . . (1)$

Using energy conservation,

$m g ℓ (1 - \cos 60 °) = \frac{1}{2} (m ℓ^{2}) ω_{m}^{2}$

$ω_{m} = \sqrt{\frac{g}{ℓ}} = 10 rad/s$

From eq.(1),

$ε_{\max} = \frac{2 \times 10 \times 0.01}{2} = 0.1 V$

$= 100 mV$

Q 19 :

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is $2 Ω$ then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is_______ N. [2026]

$5.7 \times 10^{- 2}$
$7.5 \times 10^{- 2}$
$5.7 \times 10^{- 3}$
$7.5 \times 10^{- 3}$

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(4)

To maintain constant speed

$F_{ext} = F_{B}$

$\Rightarrow F_{ext} = i l B$

$= (\frac{v B l}{R}) l B$

$= \frac{B^{2} l^{2} v}{R}$

$= \frac{{(0.1)}^{2} \times {(1)}^{2} \times 1.5}{2}$

$= 7.5 \times 10^{- 3} N$

Q 20 :

XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and the separation between them is $l$ . A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L $(L > l)$ and mass m is made to slide down from rest under gravity as shown in the figure. The terminal speed acquired by the rod is _______ m/s. (g = acceleration due to gravity) [2026]

$\frac{2 m g R}{B^{2} L^{2}}$
$\frac{m g R}{B^{2} l^{2}}$
$\frac{8 m g R}{B^{2} l^{2}}$
$\frac{2 m g R}{B^{2} l^{2}}$

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(2)

$At equilibrium (or for terminal velocity)$

$m g = I B ℓ \Rightarrow m g = (\frac{B v ℓ}{R}) B ℓ$

$V = \frac{m g R}{B^{2} ℓ^{2}}$

Topic Question Set

Q 12 :

An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when it moves with a velocity of [2023]

0%

0%

0%

0%

0%

Q 19 :

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is $2 Ω$ then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is_______ N. [2026]

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