The magnetic force acting on a charged particle of charge - 2 μ C in… | NEET 2009 (AIPMT)

Q109NEET 2009 · AIPMTEasy

Moving Charge in a Magnetic Field

The magnetic force acting on a charged particle of charge $-$ 2 $\mu$ C in a magnetic field of 2 T acting in y direction, when the particle velocity is $\left(2\hat{i} + 3\hat{j}\right) \times 10^{6}\ ms^{-1}$ , is

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Answer: (d)

$\vec{F} = q(\vec{v} \times \vec{B})$ with $\vec{B} = 2\hat{j}$ T.
$\vec{v} \times \vec{B} = (2\hat{i} + 3\hat{j}) \times 10^6 \times 2\hat{j} = 4 \times 10^6 \hat{k}$ (since $\hat{j} \times \hat{j} = 0$ ).
$\vec{F} = -2 \times 10^{-6} \times 4 \times 10^6 \hat{k} = -8\hat{k}$ N
So the answer is (d) 8 N in $-z$ direction.

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