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QJEEM24Ph9ff5: JEE Main 2024 Physics - Class 12 JEEM24Ph9ff5 Solution: A horizontal straight wire long extending from east to west…

A horizontal straight wire 5 m5 \mathrm{~m} long extending from east to west falling freely at right angle to horizontal component of earths magnetic field 0.60×10−4 Wbm−20.60 \times 10^{-4} \mathrm{~Wbm}^{-2} . The instantaneous value of emf induced in the wire when its velocity is 10 ms−110 \mathrm{~ms}^{-1} is _________ ×10−3 V\times 10^{-3} \mathrm{~V} .

Step-by-Step Verified Solution

amp;BH=0.60×10−4 Wb/m2amp; Induced emf e=BHvℓamp;=0.60×10−4×10×5amp;=3×10−3 V\begin{aligned}& \mathrm{B}_{\mathrm{H}}=0.60 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \\& \text { Induced emf e}=\mathrm{B}_{\mathrm{H}} \mathrm{v} \ell \\&=0.60 \times 10^{-4} \times 10 \times 5 \\&=3 \times 10^{-3} \mathrm{~V}\end{aligned}

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