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A bar magnet having a magnetic moment of `1.0 xx 10^4 J T^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B = 4 xx 10^(-5) T` exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction `60^@` from the field.A. `0.2J`B. `2.0J`C. `4.18J`D. `2xx10^(2)J` |
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Answer» Correct Answer - A Magnetic moment of bar `M=10^(4)J//T` `B=mu_(0)/(4pi).(2sqrt(2)M)/d^(3)` Hence, work done `W=vec(M).vec(B)` `=10^(4)xx4xx10^(-5)xxcos 60^(@) =0.2J` |
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