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When the radius of the circular current carrying coil is doubled and the current in it is halved, then the new magnetic dipole moments of the coil, if its initial value is 4 units is

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2 units

4 units

8 units

16 units

answer is C.

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Detailed Solution

$\begin{array}{l} M = {NiπR}^{2}; {MαiR}^{2} \\ \frac{M_{1}}{M_{2}} = (\frac{i_{1}}{i_{2}}) {(\frac{R_{1}}{R_{2}})}^{2} = \frac{2}{1} \times {(\frac{1}{2})}^{2}; \frac{4}{M_{2}} = \frac{1}{2}; M_{2} = 8 \end{array}$

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