Top Electromagnetic Induction PYQs for NEET 2025 Preparation

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Topic Question Set

Q 1 :
A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is [2019]

0.25 V
0.125 V
0.5 V
zero

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(2)

Here, $B = 0.1$ $T, r = 0.5$ $m, ω = 10$ $rad/s$

So, the emf generated between its centre and rim is,

$ε = \frac{1}{2} B ω r^{2} = \frac{1}{2} \times 0.1 \times 10 \times {(0.5)}^{2} = 0.125$ $V$

Q 2 :
A conducting square frame of side $' a'$ and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $' V'$ . The emf induced in the frame will be proportional to [2015]

[IMAGE 247]

$\frac{1}{{(2 x + a)}^{2}}$
$\frac{1}{(2 x - a) (2 x + a)}$
$\frac{1}{x^{2}}$
$\frac{1}{{(2 x - a)}^{2}}$

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(2)

[IMAGE 248]

Here, $P Q = R S = P R = Q S = a$

Emf induced in the frame, $ε = B_{1} (PQ) V - B_{2} (RS) V$

$= \frac{μ_{0} I}{2 π (x - a / 2)} a V - \frac{μ_{0} I}{2 π (x + a / 2)} a V$

$= \frac{μ_{0} I}{2 π} [\frac{2}{(2 x - a)} - \frac{2}{(2 x + a)}] a V$

$= \frac{μ_{0} I}{2 π} \times 2 [\frac{2 a}{(2 x - a) (2 x + a)}] a V$

$∴$ $ε \propto \frac{1}{(2 x - a) (2 x + a)}$

Q 3 :
A thin semicircular conducting ring (PQR) of radius $r$ is falling with its plane vertical in a horizontal magnetic field $B$ , as shown in the figure. The potential difference developed across the ring when its speed is $v$ , is [2014]

[IMAGE 249]

zero
$\frac{B v π r^{2}}{2}$ and $P$ is at higher potential
$π r B v$ and $R$ is at higher potential
$2 r B v$ and $R$ is at higher potential

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(4)

Motional emf induced in the semicircular ring PQR is equivalent to the motional emf induced in the imaginary conductor PR.

i.e., $ε_{P Q R} = ε_{P R} = B v l = B v (2 r)$ $(l = P R = 2 r)$

Therefore, potential difference developed across the ring is $2 r B v$ with $R$ at higher potential.

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