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Question

If the magnetic moment of magnet is 0.8×103 Am2 and the force acting on each pole in a uniform magnetic field strength of 0.48 gauss, is 0.512×106N . The distance between the poles of the magnet is : (in cm)


A
25
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B
12.5
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C
15
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D
7.5
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Solution

The correct option is D 7.5
The relation between Torque and force is τ=μB
τ=Fr
where r is distance Fr=μB
putting values of F=0.512×106N,B=0.48gauss,μ=0.8×103Am2
r=μBF=0.480.8×103Am20.512106=0.075m=7.5cm

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