Question

A wire forms a halfcircle of radius R = 70cm and two (radial) straight segments each of length L = 54cm. The wire carries current i = 21mA. What is the magnitude of the net magnetic field at the semicircle's center in terms of uT? Note: The constant μ_o is equal to 4 Pi x 10^-7 Tm/A

Answer 1

Given Data

radius = 70 cm = 0.7 m
Length of the segment , L = 54 cm
current, i = 21 mA = 21*10^-3 A

Solution :-

The net magnetic field B = B1+ B2 + B3

but B1 = B3 = 0 (because it has no magnetic field)

B = B1 + B2 + B3

   = 0 + (μ0/4π)(iπ/R) + 0

   = μ0*i / 4*R

= (1*10^-7)*(21*10^-3)*π / 0.7

= 9.42*10^-9 T

= 0.00942 μT