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A length $L$ of wire carries a current $i$ . Show that if the wire is formed into a circular coil, then the maximum torque in a given magnetic field is developed when the coil has one turn only, and that maximum torque has the magnitude $τ = \frac{L^{2} i B}{4 π}$ .

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$τ_{m a x} = \frac{i B L}{4 π}$

$τ_{m a x} = \frac{i B^{2} L}{4 π}$

$τ_{m a x} = \frac{i B L^{2}}{4 π}$

$τ_{m a x} = \frac{i B L^{2}}{2 π}$

answer is D.

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Detailed Solution

$∵ L = 2 πR \Rightarrow R = \frac{L}{2 π} M = iA = i ({πR}^{2}) = (i) (π) {(\frac{L}{2 π})}^{2} = \frac{{iL}^{2}}{4 π} τ_{\max} = MBsin 90 ° = \frac{{iBL}^{2}}{4 π}$

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