Question

A bar magnet has poles of strength 40 Am, the poles being 13 cm apart. Find the magnetic induction at a point which is 12 cm from the N-pole and 5 cm from the S-pole.

Answer 1

Due to N pole , at point x of the magnetic induction

$B_1=\frac{\mu_0}{4\pi}\frac{q_m}{l_1^2}=\frac{4 \pi *10^{-7}}{4\pi}\frac{40}{0.12^2}=2.78*10^{-4}T$

Due to S pole , at point x of the magnetic induction

$B_2=\frac{\mu_0}{4\pi}\frac{q_m}{l_2^2}=\frac{4 \pi *10^{-7}}{4\pi}\frac{40}{0.05^2}=1.6*10^{-3}T$