Question 360: Electromagnetic Induction

Question

A coil 10 turns and a resistance of $$40 \Omega$$ is connected in series with B.G. of resistance $$30 \Omega$$. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction $$10^{-2} \mathrm{~T}$$. If it is now turned through an angle of $$60^{\circ}$$ about on axis in its plane. Find the charge indicted in the coil. (Area of a coil $$=10^{-2} \mathrm{~m}^2$$ )

Accepted Answer

Given, $$n=10$$ turns
$$R_{	ext {coil }}=40 \Omega, R_G=-30 \Omega$$
total resistance in the circuit $$=40+30=70 \Omega$$
$$A=10^{-2} \mathrm{~m}^2, B=10^{-2} \mathrm{~T} 	ext {, and } 	heta_1=0, 	heta=60^{\circ}$$
we know that, flux of magnetic field is $$\phi$$
$$\phi=n B A$$
By Faraday's law, $$e=-\frac{d \phi}{d t} \quad(\because e=$$ induced emf$$)$$
$$\begin{aligned}
i= & \frac{e}{R} \Rightarrow \frac{d q}{d t}=\frac{d \phi}{R d t} \Rightarrow q=\frac{d \phi}{R} \
q & =\frac{\phi_1-\phi_2}{R}=\frac{B n A \cos 	heta_1-B n A \cos 	heta_2}{R} \
& =\frac{B n A\left(\cos 0^{\circ}-\cos 60^{\circ}ight)}{R} \
q & =\frac{B n A(1-0.5)}{R}=\frac{0.5 	imes 10^{-2} 	imes 10 	imes 10^{-2}}{70} \
& =\frac{50 	imes 10^{-5}}{70} \
\end{aligned}$$
$$q=0.714 	imes 10^{-5} \mathrm{C}$$. (Charge induced in a coil)

Answer

$$2 	imes 10^{-5} \mathrm{C}$$

Answer

$$3.2 	imes 10^{-5} \mathrm{C}$$

Answer

$$0.714 	imes 10^{-5} \mathrm{C}$$

Answer

$$55 	imes 10^{-5} \mathrm{C}$$

Electromagnetic Induction

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