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

Answer: c) 1.5 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