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  • If the bar magnet in exercise 5.13 is turned around by 180, where will the new null points be located?

5.14. If the bar magnet in exercise 5.13 is turned around by 180 \degree, where will the new null points be located?

 

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Given, d = 14 cm

The magnetic field at a distance d from the centre of the magnet on its axis:

B = \mu_{0}m/2\pi d^{3}

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial (perpendicular bisector) line.

The magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

B = \mu_{0}m/4\pi d'^3

Equating these two, we get:

\\ \frac{1}{2d^3} = \frac{1}{4d\ '^3}\\ \\ \implies \frac{d\ '^3}{d^3} = \frac{1}{2}

d' = 14 x 0.794 = 11.1cm

The new null points will be at a distance of 11.1 cm on the normal bisector.

Posted by

HARSH KANKARIA

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