Q. 254.0( 10 Votes )

# 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 :

Given:

Number of turns, N = 20

Radius of coil, r = 10cm

Magnetic field strength = 0.10 T

Current in the coil = 5 A

a) The torque on the coil is zero because the magnetic field is uniform along the coil.

b) The total force on the coil is Zero due to uniform magnetic field.

c) Let average force on each electron in the coil due to magnetic field = F

Cross-sectional area of the wire, A = 10^{-5} m^{2}

Free electron density or number of free electron per metre cube of copper, D = 10^{29}

Charge in each electron = 1.6 × 10^{19} C

We know that,

…(1)

Where,

F = force due to magnetic field

B = Magnetic field strength

e = charge on electron

V_{d} = drift velocity of electron

V_{d} = I/ (NeA) …(2)

Where, I = current through the coil

N = Density of free electron

A = cross sectional area.

Putting values in equation (2)

⇒

⇒ V_{d} = 3.125 × 10^{43} ms^{-1}

F = B × e × V_{d} …(3)

By putting the values of B, e and V_{d} in equation 3, we get,

⇒

⇒ F = 5 × 10^{-25} N

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

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