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Q.
Magnetic field due to a bar magnet 2 cm long having a pole strength of 100 Am at a point 10
cm from each pole is
Solution:
Magnetic field at a point on equatorial line of a bar magnet is given by, B = $\frac{\mu_0}{4 \pi} \frac{M}{(r^2 + l^2)^{3/2}}$
Here, M = $m \times 2l = ( 100 \times 2 \times 10^{-2})$
and $(r^2 + l^2)^{1/2} = (10 \times 10^{-2})$
$\therefore $ B = $\frac{10^{-7} \times 100 \times 2 \times 10^{-2}}{(10 \times 10^{-2})^3} = 2 \times 10^{-4}$ T