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A circular coil with cross - sectional area 0.1 cm^(2) is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field . Calculate (a) total torque on the coil (b) total force on the coil (c ) average force on each electron in the coil due to the magnetic field of the free electron density for the material of the wire is 10^(28) m^(-3) . |
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Answer» Solution :Cross sectional area of coil, A = 0.1 `cm^(2)` A = `0.1 xx 10^(-4) m^(2)` Uniform magnetic field of strength , B = 0.2 T CURRENT passing in the coil, I = 3A angle between the magnetic field and normal to the coil, `theta = 0^(@)` (a) Total TORQUE on the coil, `tau` = ABI sin `theta = 0.1 xx 10^(-4) xx 0.2 xx 3 sin 0^(@) sin 0^(@) = 0 ` `tau` = 0 (b) Total force on the coil F = BIL sin `theta = 0.2 xx 3 xx l xx sin 0^(@)` Average force: `F = q V_(d) B)` Drift velocity , `V_(d) = (I)/("ne A")"" [ because q = e ]` F = e `((I)/("ne A"))B ""[ because n = 10^(28) m^(-3)]` = `(IB)/(nA) = (3 xx 0.2)/(10^(28) xx 0.1 xx 10^(-4) ) = 6 xx 10^(-24)` `F_(av) = 0.6 xx 10^(-23) `N |
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