Question

This question has multiple correct options

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Updated on : 2022-09-05

Solution

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Correct options are C) and D)

$≃71 (3i^−5j^ +4k^)$

$∴$ area $A=(3i^−5j^ +4k^)m_{2}$

Magnetic field $B=(0.2i^+0.2j^ −0.3k^)T$

Current $I=2A$

$∴$ magnetic moment $M=IA=(6i^−10j^ +8k^)Am_{2}$

$∣M∣≃14Am_{2}(2×7=14)$

Potential energy U$=−M.B$

$=−[1.2−2−2.4]=3.2J$

Let $θ$ be the angle between $n^$ and $B$(i.e.,)between $M$ and $B.$

Now,U$=−MBcosθ.$

$∴cosθ=−MBU =−14×0.43.2 (∵0.17 ≃0.4)$

$=−74 =−0.57$

$θ=π−cos_{−1}(0.57)$

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