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A metallic rod of length $1 \mathrm{~m}$ held along east-west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_H=3 \times 10^{-5} \mathrm{~T}$. the emf induced in the rod at an instant $t=2 \mathrm{~s}$ after it is released is
(Take, $g=10 \mathrm{~ms}^{-2}$ )
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(Take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$3 \times 10^{-4} \mathrm{~V}$
Given, $B_H=3 \times 10^{-5} \mathrm{~T}$
$l=1 \mathrm{~m}$
Height travelled by rod in $t=2 \mathrm{~s}$
$h=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 2^2=20 \mathrm{~m}$
$\therefore$ Speed of rod, $v=\frac{h}{t}=\frac{20}{2}=10 \mathrm{~m} / \mathrm{s}$
Hence, induced emf
$\begin{aligned} e & =B_H v l=3 \times 10^{-5} \times 10 \times 1 \\ & =3 \times 10^{-4} \mathrm{~V}\end{aligned}$
$l=1 \mathrm{~m}$
Height travelled by rod in $t=2 \mathrm{~s}$
$h=\frac{1}{2} g t^2=\frac{1}{2} \times 10 \times 2^2=20 \mathrm{~m}$
$\therefore$ Speed of rod, $v=\frac{h}{t}=\frac{20}{2}=10 \mathrm{~m} / \mathrm{s}$
Hence, induced emf
$\begin{aligned} e & =B_H v l=3 \times 10^{-5} \times 10 \times 1 \\ & =3 \times 10^{-4} \mathrm{~V}\end{aligned}$
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