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Question: Answered & Verified by Expert
A circular coil of radius $10 \mathrm{~cm}, 500$ turns and resistance $2 \Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.25 \mathrm{~s}$. The current induced in the coil is (Horizontal component of the earth's magnetic field at that place is $3.0 \times 10^{-5} \mathrm{~T}$ )
PhysicsElectromagnetic InductionAIIMSAIIMS 2017
Options:
  • A $1.9 \times 10^{-3} \mathrm{~A}$
  • B $2.9 \times 10^{-3} \mathrm{~A}$
  • C $3.9 \times 10^{-3} \mathrm{~A}$
  • D $4.9 \times 10^{-3} \mathrm{~A}$
Solution:
1174 Upvotes Verified Answer
The correct answer is: $1.9 \times 10^{-3} \mathrm{~A}$
Initial magnetic flux through the coil,
$\begin{aligned} \phi_i & =B_H A \cos \theta=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 0^{\circ} \\ & =3 \pi \times 10^{-7} \mathrm{~Wb}\end{aligned}$
Final magnetic flux after the rotation
$\begin{aligned} \phi_f & =3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 180^{\circ} \\ & =-3 \pi \times 10^{-7} \mathrm{~Wb}\end{aligned}$
Induced emf, $\varepsilon=-N \frac{d \phi}{d t}=-\frac{N\left(\phi_f-\phi_i\right)}{t}$
$\begin{aligned} & =-\frac{500\left(-3 \pi \times 10^{-2}-3 \pi \times 10^{-7}\right)}{0.25} \\ & =\frac{500 \times\left(6 \pi \times 10^{-7}\right)}{0.25}=3.8 \times 10^{-3} \mathrm{~V} \\ & I=\frac{\varepsilon}{R}=\frac{3.8 \times 10^{-3} \mathrm{~V}}{2 \Omega}=1.9 \times 10^{-3} \mathrm{~A}\end{aligned}$

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