Question 303: Electromagnetic Induction

Question

A conducting wire of length $$2500 \mathrm{~m}$$ is kept in east-west direction, at a height of $$10 \mathrm{~m}$$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $$=25 \sqrt{2} \Omega$$, acceleration due to gravity $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{~B}_{\mathrm{H}}=2 	imes 10^{-5} \mathrm{~T}$$ )

Accepted Answer

$$\begin{aligned}
& 	ext { We know, } \mathrm{e}=\mathrm{B} / \mathrm{v} \
& \quad v^2=2 g h \
& 	herefore \quad \mathrm{v}=\sqrt{2 \mathrm{gh}} \
& 	herefore \quad \mathrm{e}=\left(2 	imes 10^{-5}ight) 	imes 2500 	imes \sqrt{2 	imes 10 	imes 10} \
& =\left(2 	imes 10^{-5}ight) 	imes 25000 	imes \sqrt{2} \
& =2 \sqrt{2} 	imes 25 	imes 10^3 	imes 10^{-5} \
& =50 \sqrt{2} 	imes 10^{-2} \mathrm{~V} \
& 	herefore \quad I=\frac{\mathrm{e}}{\mathrm{R}}=\frac{50 \sqrt{2} 	imes 10^{-2}}{25 \sqrt{2}}=2 	imes 10^{-2}=0.02 \mathrm{~A} \
\end{aligned}$$

Answer

$$0.2 \mathrm{~A}$$

Answer

$$0.02 \mathrm{~A}$$

Answer

$$0.01 \mathrm{~A}$$

Answer

$$2 \mathrm{~A}$$

Electromagnetic Induction

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