Physics · JEE | Moving Charges and Magnetism
हिन्दीAn electron moves with velocity \( v = 3 \times 10^6 \text{ m/s} \) along the \( x \)-axis in a uniform magnetic field \( B = (2 \hat{j} + 3 \hat{k}) \text{ T} \). The magnitude of the magnetic force acting on the electron is approximately:
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Correct answer · Option B
\( 2.4 \times 10^{-12} \text{ N} \)
Explanation
Force \( \vec{F} = q(\vec{v} \times \vec{B}) = -e [ (v\hat{i}) \times (2\hat{j} + 3\hat{k}) ] = -e v (2\hat{k} - 3\hat{j}) \). Magnitude \( F = e v \sqrt{2^2 + 3^2} = (1.6 \times 10^{-19})(3 \times 10^6)\sqrt{13} \approx 2.4 \times 10^{-12} \text{ N} \).