If a circular coil of 100 turns and radius 10 cm carries a current of 1 A, its magnetic dipole moment is

4 π A m^{2}

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3 π A m^{2}

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2 π A m^{2}

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1 π A m^{2}

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Solution

The correct option is D $1 π A m^{2}$
Magnetic dipole moment=n.i.A
= $100 \times 1 \times π \times (.1)^{2}$
= $1. π . A m^{2}$

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If a circular coil of 100 turns and radius 10 cm carries a current of 1 A, its magnetic dipole moment is

The correct option is D 1πAm2Magnetic dipole moment=n.i.A=100×1×π×(.1)2=1.π.Am2

The correct option is D $1 π A m^{2}$
Magnetic dipole moment=n.i.A
= $100 \times 1 \times π \times (.1)^{2}$
= $1. π . A m^{2}$