Five very long insulated straight wires are bound together to form a small cable. Currentys carried by the wires are `I_1=20 A, I_2=-5A, I_3=10 A, I_4=+7 A and I_5=-12 A`. What is the magnetic induction at a perpendicular distance of 5 cm from the cable?

Options

A. `60muT`
B. `70muT`
C. `75muT`
D. `80muT`

Answer

D

Explanation

The net current carried by the cable wires, `i_("net")=(20-5+10+7-12)A=20 A therefore B=(mu_0)/(4pi)((2i)/r)=10^(-7) xx (2 xx 20)/(5 xx 10^(-2)) =8 xx 10^(-5)=80 xx 10^(-6)T therefore B=80muT`