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CBSE Class 12-science Answered

A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate 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 Assume the area of cross-section of the wire to be 10-5m2 and the free electron density is 1029/m3.
Asked by Topperlearning User | 28 Apr, 2015, 01:54: PM
answered-by-expert Expert Answer

Here n = 200, r = 10 cm = 0.1 m

B = 0.5 T, I = 3 A

(a) Torque is given by the relation:

begin mathsize 11px style straight tau equals NIBA space sin space straight alpha space end style

begin mathsize 11px style therefore straight tau space equals space 200 space cross times space 3 space cross times space 0.5 space cross times space straight pi space cross times left parenthesis 0.1 right parenthesis squared space sin space 0 to the power of 0 end style

begin mathsize 11px style rightwards double arrow space straight tau space equals 0 end style

(b) The total force on a current loop placed in a magnetic field is always zero.

(c) Given N  = 1029/m3, A = 10-5m2

Average force on an electron of charge (e), moving with drift velocity (Vd) in the magnetic field (B), is given by:

F = BeVd

begin mathsize 11px style straight F space equals space Be straight I over neA left square bracket because space straight I space equals space neAV subscript straight d right square bracket straight F space equals space BI over NA space equals space fraction numerator 0.5 space straight x space 3 over denominator 10 to the power of 29 space straight x space 10 to the power of negative 5 end exponent end fraction space space space equals space 1.5 space straight x space 10 to the power of negative 24 end exponent space straight N end style

Answered by | 28 Apr, 2015, 03:54: PM

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