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, 21.5 cm, 539 turns, and 5.33 A. For the inner solenoid the corresponding quantities are 18.1 cm, 395 turns, and 1.95 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 5.95 mm?

Answer 1

Answer:

6700 m/s

Explanation:

The magnetic field due to a solenoid is given by B = μ₀in where i = current, n = number of turns per unit length = N/l and μ₀ = 4π × 10⁻⁷ H/m

Let B₁ be the magnetic field for the outer solenoid. For this solenoid, i = 5.33 A, n = N/l = 539 turns/0.215 m since l = 21.5 cm = 0.215 m

B₁ = 5.33 A × 539 turns/0.215 m × 4π × 10⁻⁷ H/m = 0.017 T

Let B₂ be the magnetic field for the inner solenoid. For this solenoid, i = 1.95 A, n = N/l = 395 turns/0.181 m since l = 18.1 cm = 0.181 m

B₂ = 1.95 A × 395 turns/0.181 m × 4π × 10⁻⁷ H/m = 0.0053 T

Since the magnetic fields are in opposite direction, the net magnetic field is B = B₁ - B₂ = 0.017 T - 0.0053 T = 0.0117 T.

This magnetic field produces a magnetic force on the proton which is equal to the centripetal force on the proton. So at r = 5.95 mm

Bev = mv²/r

v = Ber/m = 0.0117 T × 1.6 × 10 ⁻¹⁹ C × 5.95 × 10⁻³ m/1.67 10⁻²⁷ kg = 6669.7 m/s ≅ 6700 m/s