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Question -

A circular coil of 20 turns andradius 10 cm is placed in a uniform magnetic field of 0.10 T normal to theplane of the coil. If the current in the coil is 5.0 A, what is the

(a) totaltorque on the coil,

(b) totalforce on the coil,

(c) averageforce on each electron in the coil due to the magnetic field?

(The coil is made of copper wireof cross-sectional area 10−5 m2, and the freeelectron density in copper is given to be about 1029 m−3.)



Answer -

Number of turns on the circularcoil, 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 totaltorque on the coil is zero because the field is uniform.

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

(c) Cross-sectionalarea of copper coil, A = 10−5 m2

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

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

Magnetic force, F = Bevd

Where, 

vd = Driftvelocity of electrons

Hence, the average force on each electron is 

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