Question

Two short bar magnets each of moment '$M$' are arranged as shown in figure. The magnetic induction at point '$p$' is:

• A. $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\left(\frac{\sqrt{3}}{2}\right)$
• B. $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}(2\sqrt{2}+1)$
• C. $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\left(\frac{2\sqrt{3}+1}{3}\right)$
• D. $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}(\sqrt{3}+1)$

Answer 1

The correct option is A $\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\left(\frac{\sqrt{3}}{2}\right)$

Consider magnetic field due to left bar magnet

Resolving bar magnetic moment along and perpendicular to line joining $P$ and bar magnet

Along line joining $P$ and bar magnet:

$B_1=\frac{2{\mu }_{0}M}{4\pi d^3}\cos {60}^o$

Similarly we can write for right bar magnet the magnetic field components, their $\cos$ components add up and $\sin$ components cancel each other.

$B_{1_{total}}=\frac{2{\mu }_{0}M\cos {60}^o}{4\pi d^3}\times 2\cos {30}^o=\frac{\sqrt{3}{\mu }_{o}M}{4\pi d^3}$

Perpendicular to line joining P and bar magnet:

$B_2=\frac{{\mu }_{0}M\sin {60}^o}{4\pi d^3}\times 2\sin {30}^o$

Similarly we can write for right bar magnet the magnetic field components,their $\sin$ components add up and $\cos$ components cancel each other.

$B_{2_{total}}=\frac{{\mu }_{0}M\sin {60}^o}{4\pi d^3}$

$B_{total}=\frac{{\mu }_{0}M}{4\pi d^3}\frac{\sqrt{3}}{2}$