A wire carrying a current I along the positive x -axis has length L. It is kept in a magnetic field B = (2i + 3j - 4k) T. The magnitude of the magnetic force acting on the wire is: (a) √(5) IL (b) 5 IL (c) √(3) IL (d) 3 IL

Answer: (b). F = IL × B with L = Li; the i part of B gives no force. i × (2i + 3j - 4k) = 3k + 4j |F| = IL√(3² + 4²) = 5 IL So the answer is (b).

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Q24NEET 2023Easy

Force on Current-Carrying Conductors

A wire carrying a current $I$ along the positive $x$ -axis has length $L$ . It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})\text{ T}$ . The magnitude of the magnetic force acting on the wire is:

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Answer: (b)

$\vec{F} = I\vec{L} \times \vec{B}$ with $\vec{L} = L\hat{i}$ ; the $\hat{i}$ part of $\vec{B}$ gives no force.
$\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k}) = 3\hat{k} + 4\hat{j}$
$|\vec{F}| = IL\sqrt{3^2 + 4^2} = 5\,IL$
So the answer is (b).

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