Question 25

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in 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 magnetic field?

(The coil is made of copper wire of cross-sectional area 10^{-5} m^{2}, and the free electron density in copper is given to be about 10^{29} m^{-3}.)

Answer

Number of turns on the circular coil, n = 20

Radius of the coil,r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

(a) The total torque on the coil is zero because the field is uniform.

(b) The total force on the coil is zero because the field is uniform.

(c) Cross-sectional area of copper coil, A = 10^{-5} m^{2}

Number of free electrons per cubic meter in copper, N = 10^{29} /m^{3}

Charge on the electron, e = 1.6 × 10^{-19} C

Magnetic force, F = Bev_{d}

Where,

v_{d }= Drift velocity of electrons

Hence, the average force on each electron is 5 x 10^{-25} N.

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Madanuru Thridhamni
2019-08-13 20:42:13

Here how N is 10^29 given N=20 turns

- NCERT Chapter

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