Question

A short bar magnet has a magnetic moment of \[0.39J{T^{ - 1}}\]. The magnitude and direction of the magnetic field produced by the magnet at a distance of 20cm from the center of the magnet on the equatorial line of the magnet is
A. 0.049 G, N-S direction
B. 4.9 G, S-N direction
C. 0.0195 G, S-N direction
D. 19.5 G, N-S direction

Answer 1

Hint: For addressing these types of problems you will always recall the magnetic field equations, how to measure the magnetic field and how to locate the direction of the measured magnetic field.

Complete Step-by-Step solution:
According to the question we have to find out the magnitude and direction of the magnetic field produced by the short bar magnet. We have given magnetic moment = 0.39JT and diameter is given as 20cm so the radius will be 0.2m.On the equatorial line of magnet
Using the magnetic field formula to find the direction
$B = \dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{m}{d}$ (Equation 1)
Substituting the values in equation 1
$ \Rightarrow $$B = {10^{ - 7}} \times \dfrac{{0.39}}{{{{(0.2)}^3}}}$
$ \Rightarrow $$B = \dfrac{{0.39}}{8} \times {10^{ - 4}}$
$ \Rightarrow $$0.049 \times {10^{ - 4}}T,N - S$ Direction
$ \Rightarrow $0.049 G; N-S direction
Hence, option A is correct.

NOTE: A shallow bar magnet aligns its pole in the north-south orientation of the compass in a horizontal plane. Null points tend to be found on the magnet pole at 14 cm from the middle of the magnet. Hence the magnetic field in the direction of the earth’s magnetic field is 0.54G.