Question

Two short bar magnets are arranged as shown. Find resultant magnetic induction at the point '$P$'. Also, find resultant magnetic induction at the same point '$P$' if poles of a magnet of higher magnetic moment are reversed.

• A. $\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{8},\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{20}$
• B. $\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{8},\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{40}$
• C. $\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{5},\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{40}$
• D. $\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{4},\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{40}$

Answer 1

The correct option is B $\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{8},\frac{{\mu }_{0}M}{4\pi d^3}\sqrt{40}$

$B_1=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$ ..(1)

$B_2=\frac{{\mu }_0}{4\pi }\frac{2m}{d^3}$ ..(2)

$B_3=\frac{{\mu }_0}{4\pi }\frac{2\left(2m\right)}{d^3}$ ..(3)

In this situtaion resultant feild $B$ is,

$B=\sqrt{B^2+{(B_3-B_1)}^2}$

$=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\sqrt{2^2+{(4-2)}^2}$

$=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\left(\sqrt{8}\right)$

Case 2 when magnet 3 will reversed.

The direction of $B_3$ is change from right to left then resultant field is $B^{\prime }$ ,

$B^{\prime }=\sqrt{{B_2}^2+(B_1+B_3{)}^2}$

$=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\sqrt{2^2+{(2+4)}^2}$

$=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}\sqrt{40}$