Question

Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?

Answer 1

Answer:

$I_2=6.8A$

Explanation:

From the question we are told that:

Turns of inner coil $N_1=120$

Radius of inner coil $r_1=0.012m$

Current of  inner coil $I_1=6.0A$

Turns of Outer coil $N_2=150$

Radius of Outer coil $r_2=0.017m$

Generally the equation for Magnetic Field is mathematically given by

$B =\frac{ \mu N I}{2R}$

Therefore

Condition for the net Magnetic field to be zero

$\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}$

$I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}$

$I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}$

$I_2=6.8A$