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# A circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.1 T normal to the plane of the coil. If the current in the coil is 5.0 A What is the 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 to power  minus five m square and the free electron density in copper is given to be about 10 power 29 m to power minus three.

## 1 Answers

2 years ago

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

= I / NeA

F = 5 * 10^-25 N

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