Magnetic Effects & Magnetism

JEE Main level, single correct, medium

259 of 276

Question 259 of 276

A square loop is carrying a steady current II and the magnitude of its magnetic dipole moment is mm . If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

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Answer: (A) 4mπ\dfrac{4m}{\pi }

Let the side of square loop be aa.
The magnetic dipole is∴m=Ia2∴m=I{a}^{2}
a=mIa=\sqrt{\dfrac{m}{I}}
Radius of circular loop rr related to aa as circumference of circular loop is equal to perimeter of square loop.
2πr=4a=4mI2\pi r=4a=4\sqrt{\dfrac{m}{I}}
r=2πmIr=\dfrac{2}{\pi }\sqrt{\dfrac{m}{I}}
∴∴ New dipole moment m′=π(2πmI)2I{m}^{'}=\pi {\left(\dfrac{2}{\pi }\sqrt{\dfrac{m}{I}}\right)}^{2}I
=4mπ=\dfrac{4m}{\pi }

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