Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is . ( 0 = 4 10-7 T m/A) [figure]
- 6 10-6 N towards P
- 6 10-6 N towards R
- 6 10-7 N towards P
- 6 10-7 N towards R
Answer
(2) 6 10-6 N towards R
Solution
Force between two long parallel wires: F = 0 I1 I2 l2 d, with l = 0.15 m.
Force on Q due to P (IP = 3 A upward, IQ = 1 A downward, d = 0.03 m):
FQP = 2 10-7 3 1 0.150.03 = 3 10-6 N
The currents are antiparallel, Q is repelled from P, i.e. pushed towards R.
Force on Q due to R (IR = 2 A downward, d = 0.02 m):
FQR = 2 10-7 2 1 0.150.02 = 3 10-6 N
The currents are parallel, Q is attracted towards R.
Fnet = 3 10-6 + 3 10-6 = 6 10-6 N towards R
Answer: (2)