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
3IL
B
5IL
C
5IL
D
3IL
Step-by-Step Solution
The magnetic force on a current-carrying wire is F=I(L×B). Here L=Li^. 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^). The magnitude is ∣F∣=IL42+32=IL16+9=5IL.
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