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CBSE Class 12 Physics 2013 Paper

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Question : 10 of 29
Marks: +1, -0
A coil of ' NN ' turns and radius ' RR ' carries a current ' TT. It is unwound and rewound to make a square coil of side 'a' having same number of turns (N)(N). Keeping the current 'T same, find the ratio of the magnetic moments of the square coil and the circular coil.
Solution:  
Magnetic moment=M=NIA\text{Magnetic moment} = M = N I A
Mcircular=NIπR2M_{\text{circular}} = N I \pi R^2
After rewinding,2πR=4a\text{After rewinding,} 2 \pi R = 4 a
∴a=πR2\therefore a = \frac{\pi R}{2}
Msquare=NIa2M_{\text{square}} = N I a^2
Or, Msquare=NI(πR2)2M_{\text{square}} = N I \left( \frac{\pi R}{2} \right)^2
∴Msquare=NIπ2R24\therefore M_{\text{square}} = \frac{N I \pi^2 R^2}{4}
Now, Msquare=π4M_{\text{square}} = \frac{\pi}{4}
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