Question

Two bar magnet having magnetic dipole moments $4 A-m^2$ and $5 A-m^2$ are kept as shown below. The resultant dipole moment will be

• A. $\sqrt{21} A m^2$
• B. $\sqrt{41} A m^2$
• C. $\sqrt{61} A-m^2$
• D. $\sqrt{31} A-m^2$

Answer 1

Answer: (C)

Here,
$ \& M_1=4 A-m^2$
$M_2=5 A-m^2$

According to parallelogram law of vector addition, resultant dipole moment $M$
$M =\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos 60^{\circ}} $
$ =\sqrt{4^2+5^2+2 \times 4 \times 5 \times \cos 60^{\circ}}=\sqrt{61} A-m^2$