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A circular coil of 20 turns and radius 10 cm is placed in a uniformmagnetic field of 0.10 T normal to the plane of the coil. If the currentin the coil is 5.0 A, what is the (a) total torque on the coil, (b) total force on the coil, (c) average force on each electron in the coil due to the magneticfield?(The coil is made of copper wire of cross-sectional area 10–5 m2, andthe free electron density in copper is given to be about 1029 m–3.)

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Solution

Given: The radius of the circular coil is 10cm, total number of turns in the coil is 20, the magnitude of the magnetic field is 0.10T normal to the plane of the coil, the magnitude of the current is 5A, the number of electron in per cubic meter of copper is 10 29 m 3 and the cross sectional area of the copper coil is 1 0 5 m 2 .

(a)

Since, the nature of the magnetic field is uniform; therefore, the net force induced in the coil is zero.

Thus, total torque on the coil is zero.

(b)

Since, the nature of the magnetic field is uniform; therefore, the net force induced in the coil is zero.

(c)

The magnetic force is given as,

F= Bel NeA = Bl NA

Where, the magnetic field is B, the length of the coil is l, the number of turns is N and the area of cross section of the coil is A.

By substituting the values in the above equation, we get

F= 0.10×5.0 10 29 × 10 5 =5× 10 25 N

Thus, the average force on each electron in the coil due to the magnetic field the average force on each electron is 5× 10 25 N.


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