JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Electromagnetic Induction

A square shaped coil of area

70 cm^{2}

having 600 turns rotates in a magnetic field of

0.4 wbm^{- 2}

, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field.
If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at

6 0^{\circ}

with the field, will be ____ V.
(Take

π = \frac{22}{7}

)

Answer: 44

Solution

The instantaneous emf of the rotating coil is

e = N B A ω sin (ω t)

, where

ω t

is the angle between the coil's normal and the field. Here

N = 600

A = 70 \times 1 0^{- 4} m^{2}

B = 0.4

T, and

ω = \frac{2 π \times 500}{60} = \frac{1100}{21} rad s^{- 1}

with

π = \frac{22}{7}

. When the plane of the coil makes

6 0^{\circ}

with the field, its normal makes

3 0^{\circ}

with the field, so

sin (ω t) = sin 3 0^{\circ} = \frac{1}{2}

. Then

e = 600 \times 70 \times 1 0^{- 4} \times 0.4 \times \frac{1100}{21} \times \frac{1}{2} = 44

. The instantaneous emf is

44

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Electromagnetic Induction

Solution

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