Q.

Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive on expression for the force per unit length between two parallel current carrying conductors?

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Detailed Solution

Flemings left hand rule: Stretch the left hand such that the thumb, the forefinger and the middle finger are mutually parpendicular to each other. If fore finger indicates the direction of magnetic field, the middle finger current direction then the thumb indicates direction of force on the conductor.
The force on a current carrying conductor in a magnetic field:
1) Consider a straight wire of length 'l' of cross sectional area A and carrying a current 'i' placed in a uniform magnetic field induction 'B' as shown in fig.

2) Due to the potential difference, electrons move with drift velocity. 'vd'
3) We know that the force 'f' acting on charge 'q' in the magnetic field of induction B is
f = qVdB sinθ......1
4) If 'n' is no of free electrons per unit volume of the conductor, the current in it is
i = nqvdA...... (2)
5) The total force F acting on 'nlA' no of electrons in length of conductor is

F = (nlA) f = nlA. qVdBsinθ

F = nq VdA (lBsinθ)      nqVdA

F = ilBsinθ

Force between two parallel conductors carrying current :

1) Consider two long parallel conductors I & II of length l carrying current i1i2 separated by a distance r.

2) The magnetic indiction field strength due to conductor I at distance 'r' is

B1=μ02πi1r 1

3) conductor II is in 'B1'. So the force on it is F2=i2lB1 (towards first conductor)

F2=μ02πi1i2lr 2

The magnetic induction due to conductor II at a distance 'r' is

B1=μ02πi2r 3

Conductor 'I' is in 'B2'. So the force on it is F1=i1lB1 (towards second conductor)

F1=μ02πi1i2rl 4

F1=F2=μ02πi1i2rl

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