KCET · Physics · Electromagnetic Induction
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, \(g=10 \mathrm{~ms}^{-2}\) )
- A \(6 \times 10^{-4} \mathrm{~V}\)
- B \(3 \times 10^{-3} \mathrm{~V}\)
- C \(3 \times 10^{-4} \mathrm{~V}\)
- D \(6 \times 10^{-3} \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(3 \times 10^{-4} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
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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