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Question: A square loop of side 2m carrying current I<sub>0</sub> is placed in x – y plane in a magnetic field...

A square loop of side 2m carrying current I0 is placed in x – y plane in a magnetic field B = (4i^+3j^)(4\widehat{i} + 3\widehat{j}) T. Find the unit vector along the axis about which it will start rotating –

A

4j^+3i^5\frac{4\widehat{j} + 3\widehat{i}}{5}

B

−4j^−3i^5\frac{- 4\widehat{j} - 3\widehat{i}}{5}

C

−4j^+3i^5\frac{- 4\widehat{j} + 3\widehat{i}}{5}

D

−4j^−3i^5\frac{- 4\widehat{j} - 3\widehat{i}}{5}

Answer

−4j^−3i^5\frac{- 4\widehat{j} - 3\widehat{i}}{5}

Explanation

Solution

Work done by external source

Wext = q(VF – Vi), 20 = 4(V – 10); V = 15 volt.