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A bar magnet having a magnetic moment of 2.0 xx 10^(-4) J//T is free to rotate in a horizontal plane. A horizontal magnetic field B = 5 xx 10^(-5) T exists in the space. Then the work done in rotating the magnet slowls from a direction parallel to the field to a direction 60^(@) from the field is |
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Answer» 0.1 J `= (-MB costheta_(2)) - (-MBcostheta_(1))` `=-M(cos60^(@)-cos0^(@))` `=1/2MB = 1/2xx(2.0xx10^(4)J//T)(5xx10^(-5)T = 0.5J` |
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