Electromagnetic Induction · Question 22

JEE Main 2025 (Online) 29th January Morning ShiftMCQEasy📈 Topic in 6% papers+4−1

A coil of area A and N turns is rotating with angular velocity in a uniform magnetic field B about an axis perpendicular to B. Magnetic flux and induced emf across it, at an instant when B is parallel to the plane of coil, are :

= AB, = NAB
= AB, = 0
= 0, = 0
= 0, = NAB

Answer

(4) = 0, = NAB

Solution

Let = t be the angle between B and the area vector (normal to the coil).

Flux through one turn: = BA t

emf: = -N d(BA t)dt = NAB t

When B is parallel to the plane of the coil, the normal is perpendicular to B = 90

= BA 90 = 0 and = NAB 90 = NAB (maximum)

Answer: (4)

JEE Main 2025 (Online) 29th January Morning Shift

Keep practising — Electromagnetic Induction

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

e1 = -L1 dI2dt - M12 dI1dt
e1 = -L1 dI1dt + M12 dI2dt
e1 = -L1 dI1dt - M12 dI1dt
e1 = -L1 dI1dt - M12 dI2dt

JEE Main 2025 (Online) 29th January Morning Shift

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 rad s-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? ( = 3.14)

0.5024 V
0.0628 V
0.2512 V
0.1256 V

JEE Main 2025 (Online) 28th January Evening Shift

[figure] A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E tn, then value of n is .

JEE Main 2025 (Online) 28th January Evening Shift

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