. Two long straight parallel wires are  2 metres apart, perpendicular to the plane of the paper (wee figure). The wire A carries a current of 9.6 amps, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10 / 11 metre from the wire B, is zero                               Find :(i) The magnitude and direction of the current in B.(ii) The magnitude of the magnetic field of induction at the point S.(iii) The force per unit length on the wire B.

Navjyot Kalra
10 years ago
Hello Student,
. (i) The magnetic field at P due to current in wire A.
BA = μ0 / 4π 2IA / rAP = μ0 / 4π x 2 x 9.6 / (2 + 10 / 11) (Direction P to M ) … (i)
NOTE : The current in wire B should be in upward direction so as to cancel the magnetic field due to A at P. (By right hand Thumb rule)
The magnetic field at P. due to current in wire B
BB = μ0 / 4π x 2IB / (10 / 11) …(ii)
From (i) and (ii)
μ­0 / 4π x 2 x 9.6 / (2 + 10 / 11) = μ­0 / 4 π x 2IB / (10 / 11)
⇒ 9.6 x 11 / 32 = IB x 11 / 10 ⇒ IB = 96 / 32 = 3A
(ii) The dimensions given shows that
SA2 + SB2 = AB2 ⇒ ∠ASB = 90°
Magnetic field due to A at S
BSA = μ0 / 4π 2IA / rSA = μ0 / 4π x 2 x 9 .6 / 1.6 (Directed S to B)
Magnetic field due to B at S
BSB = μ0 / 4π 2IB / rSB = μ0 / 4π 2 x 3 / 1.2 (Directed S to A)
The resultant magnetic field
B = √B2SA + B2SB = μ0 / 4π √(9.6 / 0.8)2 + (3 / 06 )2
= 10-7 x 13 = 1.3 x 10-6 T
= μ0 / 4π 2IAIB / rAB
= 10-7 x 2 x 9.6 x 3 / 2 = 28.8 x 10-7 N/m
This force will be repulsive in nature.
Thanks
Navjot Kalra