JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Electromagnetic Induction

A coil of 200 turns and area

0.20 m^{2}

is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is

\frac{2 π}{β}

volt. The value of

β

is ______.

Answer: 5

Solution

The flux through the rotating coil is

ϕ = N B A cos (ω t)

, so the induced emf is

ε = - \frac{d ϕ}{d t} = N B A ω sin (ω t)

, with peak value

ε_{ma x} = N B A ω

. With

N = 200

A = 0.20 m^{2}

B = 0.01

T and

ω = 2 π \times 0.5 = π

rad/s:

ε_{ma x} = 200 \times 0.2 \times 0.01 \times π = 0.4 π = \frac{2 π}{5}

V, so

β = 5

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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