Ncert Exercises & Solutions

5.13 A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

NEET SOLUTION:

Earth's magnetic field at the given place, H=0.36 G
The magnetic field ata distance d, on the axis of the magent is given as

B_{1} = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}} = H . . . (i)

Where, Ho =PemeaDiity of free space, M=magnetic moment
The magnetic field at the same distance d, on the equatotial line of
the magnet Is given as:

B_{2} = \frac{μ_{0} M}{4 π d^{3}} = \frac{H}{2} [Using equation (i)]

Total magnetic field, B = B_{1} + B_{2}

= H + \frac{H}{2}

= 0.36 + 0.18 = 0.54 G

hence, the magnetic field is 0.54 G in the direction of earth's magnetic field.

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