A steady current of 4 A flows through a circular coil having 10 turns of radius 8 cm. The coil lies in X-Y plane with its center at the origin. The magnitude and direction of the magnetic dipole moment of the coil (with respect to plane of the loop) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0.80\,\mathrm{A\,m^{2}} \), perpendicular
The magnetic dipole moment of a coil is \( m = NIA \), directed along the normal to the plane of the coil (by the right-hand rule).
\( N = 10 \)
\( I = 4\,\mathrm{A} \)
\( r = 8\,\mathrm{cm} = 0.08\,\mathrm{m} \)
\( A = \pi r^{2} = 3.14\times(0.08)^{2} = 0.0201\,\mathrm{m^{2}} \)
\( m = NIA = 10\times4\times0.0201 = 0.804\,\mathrm{A\,m^{2}} \approx 0.80\,\mathrm{A\,m^{2}} \)
The coil lies in the X-Y plane, so m is along the Z-axis, i.e. perpendicular to the plane of the loop.