A circular coil of 20 turns and ra | Class 12 Physics Chapter Moving Charges and Magnetism, Moving Charges and Magnetism NCERT Solutions

Question:

 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 m2, and the free electron density in copper is given to be about 1029 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 m2

Number of free electrons per cubic meter in copper, N = 1029 /m3

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

Magnetic force, F = Bevd

Where,

vd = Drift velocity of electrons

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


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Comments

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

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


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 25:  A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T n....