# A Conducting Loop of Area 5.0 Cm2 Is Placed in a Magnetic Field Which Varies Sinusoidally with Time As B = B0 Sin ωT Where B0 = 0.20 T and ω = 300 S−1. - Physics

Sum

A conducting loop of area 5.0 cm2 is placed in a magnetic field which varies sinusoidally with time as B = B0 sin ωt where B0 = 0.20 T and ω = 300 s−1. The normal to the coil makes an angle of 60° with the field. Find (a) the maximum emf induced in the coil, (b) the emf induced at τ = (π/900)s and (c) the emf induced at t = (π/600) s.

#### Solution

Given:-
Area of the coil, A = 5 cm2 = 5 × 10−4 m2
The magnetic field at time t is given by
B = B0 sin ωt = 0.2 sin (300t)
Angle of the normal of the coil with the magnetic field, θ = 60°

(a) The emf induced in the coil is given by

$e = \frac{- d\theta}{dt} = \frac{d}{dt}(BA \cos \theta)$

$= \frac{d}{dt}\left[ \left( B_0 \sin \omega t \right) \times 5 \times {10}^{- 4} \times 1/2 \right]$

$= B_0 \times \frac{5}{2} \times {10}^{- 4} \frac{d}{dt}(\sin \omega t)$

$= \frac{B_0 5}{2} {10}^{- 4} \omega\left( \cos \omega t \right)$

$= \frac{0 . 2 \times 5}{2} \times 300 \times {10}^{- 4} \times \cos \omega t$

$= 15 \times {10}^{- 3} cost \omega t$

The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.

Thus, the maximum value of the induced emf is given by

$e_{max} = 15 \times {10}^{- 3} = 0 . 015 V$

(b) The induced emf at t = $\left( \frac{\pi}{900} \right) s$ is given by

e = 15 × 10−3 × cos ωt

= 15 × 10−3 × cos$\left( 300\times\frac{\pi}{900} \right)$

= 15 × 10−3 × 1/2

$= \frac{0 . 015}{2} = 0 . 0075 = 7 . 5 \times {10}^{- 3} V$

(c) The induced emf at t = $\frac{\pi}{600} s$ is given by

e = 15 × 10−3 × cos $\left( 300 \times \frac{\pi}{600} \right)$

= 15 × 10−3 × 0 = 0 V

Concept: Induced Emf and Current
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 16 Electromagnetic Induction
Q 8 | Page 306