Q1. A coil of wire has an area of 600 sq cm and 500 turns. If it carries a current of 1.5 A, its magnetic dipole moment is:
A 10 A m^2
B 15 A m^2
C 30 A m^2
D 45 A m^2 Correct
Explanation
The magnetic dipole moment of a current-carrying coil is given by M = NIA, where N is the number of turns, I is current and A is area in square metres. Here, 600 sq cm equals 600/10000 = 0.06 m^2. So M = 500 x 1.5 x 0.06 = 45 A m^2. The values 10, 15 and 30 A m^2 arise only if the area conversion, number of turns or multiplication is handled incorrectly. With the given data, 45 A m^2 is correct.