A current of 50 A is passed through a straight wire of length 6 cm, then the magnetic induction at a point 5 cm from either end of the wire is (1 gauss =10−4T)

a) 2.5 gauss

b) 1.25 gauss

c) 1.2 gauss

d) 3.0 gauss

Solution:

B = (μ0I/4πr)[sinθ1 + sinθ2]

= 10-7 x (50/(4 x 10-2)] (2sin 370)

= 10-7 x (50/(0.04)] 2(0.6)

= 10-7 x (50/(0.01)] (0.3)

= 1.5 x 10-4 T

= 1.5 gauss