Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

125 of 276

Question 125 of 276

A charge of 4.0μC4.0\mu C is moving with a velocity of 4.0×106ms−14.0\times {10}^{6}m{s}^{-1} along the positive yy-axis under a magnetic field BB of strength (2k^)T(2\hat{k})T. The force acting on the charge is xi^Nx\hat{i}N. The value of xx is ______.

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Answer: 32

Given:
q=4μC,v⃗=4×106j^ms−1q=4\mu C,\vec{v}=4\times {10}^{6}\hat{j}m{s}^{-1} and B⃗=2k^T\vec{B}=2\hat{k}T
Force acting on charged particle, F⃗=q(v⃗×B⃗)\vec{F}=q(\vec{v}\times \vec{B})
=4×10−6(4×106j^×2k^)=4\times {10}^{-6}\left(4\times {10}^{6}\hat{j}\times 2\hat{k}\right)
=4×10−6×8×106i^=4\times {10}^{-6}\times 8\times {10}^{6}\hat{i}
⇒F⃗=32i^N\Rightarrow \vec{F}=32\hat{i}N
⇒x=32\Rightarrow x=32

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