Physics

Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be $U_{i}=−M.B=−MBcos0=−MB$

Now when it is deflected by an angle $θ$ then $U_{f}=−M.B=−MBcosθ$

$ΔU=U_{i}−U_{f}$

$ΔU=−MB(1−cosθ)$

Area $A=7m_{2},$ Normal $n^=50 1 (3i^−5j^ +4k^)$

$≃71 (3i^−5j^ +4k^)$

$∴$ 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$=−M.B$

$=−[1.2−2−2.4]=3.2J$

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