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NEET PHYSICSMedium

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

A

3IL3 IL

B

5IL\sqrt{5} IL

C

5IL5 IL

D

3IL\sqrt{3} IL

Step-by-Step Solution

The magnetic force on a current-carrying wire is F=I(L×B)\vec{F} = I(\vec{L} \times \vec{B}). Here L=Li^\vec{L} = L\hat{i}. So F=I(Li^×(2i^+3j^4k^))=IL(2(i^×i^)+3(i^×j^)4(i^×k^))=IL(0+3k^4(j^))=IL(4j^+3k^)\vec{F} = I(L\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k})) = IL(2(\hat{i}\times\hat{i}) + 3(\hat{i}\times\hat{j}) - 4(\hat{i}\times\hat{k})) = IL(0 + 3\hat{k} - 4(-\hat{j})) = IL(4\hat{j} + 3\hat{k}). The magnitude is F=IL42+32=IL16+9=5IL|\vec{F}| = IL \sqrt{4^2 + 3^2} = IL \sqrt{16+9} = 5IL.

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