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Question: Two identical bar magnets with a length 10 *cm* and weight 50 *gm – weight* are arranged freely with...

Two identical bar magnets with a length 10 cm and weight 50 gm – weight are arranged freely with their like poles facing in a arranged vertical glass tube. The upper magnet hangs in the air above the lower one so that the distance between the nearest pole of the magnet is 3mm. Pole strength of the poles of each magnet will be

A

6.64 amp ×m

B

2 amp ×m

C

10.25 amp ×m

D

None of these

Answer

6.64 amp ×m

Explanation

Solution

The weight of upper magnet should be balanced by the repulsion between the two magnet

μ4πm2r2=50gmwt\therefore \frac { \mu } { 4 \pi } \cdot \frac { m ^ { 2 } } { r ^ { 2 } } = 50 \mathrm { gm } - w t

107×m2(9×106)=50×103×9.810 ^ { - 7 } \times \frac { m ^ { 2 } } { \left( 9 \times 10 ^ { - 6 } \right) } = 50 \times 10 ^ { - 3 } \times 9.8

m=6.64amp×mm = 6.64 \mathrm { amp } \times m