JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2026
- Medium
- MCQ
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity ( \(g =10 m / s ^2\) ) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travells a vertical distance of 200 m is ___________ mV .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \( 20 \sqrt{10} \)
Step-by-step Solution
Detailed explanation
\(\varepsilon= vB \ell\)
\( v=\sqrt{2gh}=\sqrt{2\times10\times200}=20\sqrt{10} \)
\( \varepsilon=(20\sqrt{10})(0.5\times10^{-4})20\)
\(=20 \sqrt{10} \times 10^{-3}=20 \sqrt{10} mV\)
\( v=\sqrt{2gh}=\sqrt{2\times10\times200}=20\sqrt{10} \)
\( \varepsilon=(20\sqrt{10})(0.5\times10^{-4})20\)
\(=20 \sqrt{10} \times 10^{-3}=20 \sqrt{10} mV\)
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