Question

ws ies separated by 22 cm carry currents in the same direction. The current in one wire the other is 45 A. Find the magnitude of the magnetic field halfway between the wires. (31 C) 8. Two similar circular coils having 500 turns each radius 20 cm are set co-axially. If the coils are separated by 20 cm and carry current in the same direction, find the magnetic field at a. b. c. Midpoint of the two coils. At 5 cm from one of them At 15 cm from the same coil.

Answer 1

The magnetic field due to a single circular coil carrying Current I , Radius a with N turns at distance z along the axis of the coil is given by

$B = (\mu _{o}NIa^2/2)[1/ (a^2+z^2)^{3/2}]\hat{z}$

Now ,if we have two coils , one at z = -a/2 and z = +a/2 , the origin being at exact mid-point of the centers of two coils

then the field at this mid point is given by

$B = (\mu _{o}NI/2a)( X+Y)$

If Put z= 0

$B = (\mu _{o}NI/a)(8/5\sqrt{5})$

N =500 turns a =0.2 m I = not given by you

$B=(I)((4\pi \times 10^{-7}Weber/Amp-m) \times 500/0.2)(8/5\sqrt{5})$

= ( I ) x 22479 x 10^(-7) Tesla = ( I ) x 2.25 x 10^(-3) Tesla

b.

Field at a distance of 5 cm

$B = (\mu _{o}NI/2a)( X+Y)$

In this case z = 0.2 m + 0.05 m = 0.25 m

= 0.122

= 0.512

$B = (\mu _{o}NI/2a)( X+Y)$ = ( I ) x [4 pi x 10^(-7) x 500 / (2 x 0.2 )] ( 0.122 +0.512 )

= ( I ) x 9958 x 10^(-7) = ( I ) x 9.96 x 10^(-4) T

c. When we want field at  0.15 m from same coil z = 0.2 + 0.15 = 0.35 m   

=0.067

= 0.244

$B = (\mu _{o}NI/2a)( X+Y)$ =  ( I ) x [4 pi x 10^(-7) x 500 / (2 x 0.2 )]( 0.067+0.244) = ( I ) x 4.88 x 10^(-4)T