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InterviewSolution
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A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the : (i) Total troque on the coil, (ii) total force on the coil, (iii) average force on each electron on each electron in the coil, due to the magnetic field. Assume, the area of cross-section of the wire to be 10^(-5) m^(2) and the free electron density as 10^(29) m^(-3). |
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Answer» Solution :Here`N = 200`, radius of coil `r = 10 cm = 0.1 m` and hence AREA of circular coil `A = pi r^2 = pi (0.1)^(2) = 0.01 pi m^2`, uniform magnetic field `B = 0.5 T, theta = 0^@` and current I = 3.0 A (i) TORQUE on the coil `tau = NAI B sin theta = 200 xx 0.01 pi xx 3.0 xx 0.5 xx sin 0^@ = 0` (ii) Total force on a current carrying coil placed in a uniform magnetic field is ALWAYS zero. (iii) Average force on each electron in the coil, `F= B.e.v_d = Be 1/(n eA.)`, where n = free electron dnesity `10^(29) m^(-3)` and A. = area of cross-section of wire = `10^(-5) m^(2)` `:. F = (0.5 xx 1.6 xx 10^(-19) xx 3.0)/(10^(29) xx 1.6 xx 10^(-19) xx 10^(-5)) = 1.5 xx 10^(-24) N`. |
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