Two solenoids are nested coaxially such that their magnetic fields point in oppo

Question

Two solenoids are nested coaxially such that their magnetic fields point in opposite directions. Treat the solenoids as ideal. The outer one has a radius of 20 mm, and the radius of the inner solenoid is 10 mm. The length, number of turns, and current of the outer solenoid are, respectively, 21.3 cm, 529 turns, and 4.49 A. For the inner solenoid the corresponding quantities are 18.1 cm, 307 turns, and 1.05 A. At what speed, v1, should a proton be traveling, inside the apparatus and perpendicular to the magnetic field, if it is to orbit the axis of the solenoids at a radius of 6.23 mm? v1= ? m/s

And at what speed, v2, for an orbital radius of 14.1 mm? v2= ? m/s

please solve for v1 and v2

Explanation / Answer

B=mu_0(N/L)I

B - magnetic field, N - no. of turns, L- length, I - current.

Magnetic field due to outer solenoid (B1) =  mu_0( 529/.213)4.49 = 11151.22 mu_0 T

Magnetic field due to inner solenoid (B2) = mu_0( 307/.181)1.05 = 1780.94 mu_0 T

superposition of magnetic fields
Bnet = ( B1 - B2 ) = 9370.28mu_0

The force the proton feels is determined by the Lorentz force (in this case, F=q*V1xB, where F is force, q is positive for proton, and V is velocity).This Lorentz force will cause the proton to travel in a circle. Because of centripetal acceleration,

F=ma=m*V1^2/r

so  q*V1xB = m*V1^2/r

1.6 * 10-19 x V1 x 9370.28mu_0 = 1.672 * 10-27 x V1^2 / 6.23 * 10-3   

V1 = 1.6 * 10-19 x 9370.28 x 4pi * 10-7 x  6.23 * 10-3 / 1.672 * 10-27

V1 = 7024.04 m/s

so , When the orbital radius is 14.1mm

V2=q*B*r/m

V2 = 1.6 * 10-19 x  11151.22 x 4pi * 10-7 x 14.1 / 1.672 * 10-27

V2= 8359.95 m/s

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