Question

A solenoid with 35 turns per centimeter carries a current I. An electron moves within the solenoid in a circle that has a radius of 3.0 cm and is perpendicular to the axis of the solenoid. If the speed of the electron is 3.0 ✕ 105 m/s, what is I (in A)?

Answer 1

Answer:

The current of the solenoid is 0.0129 A.

Explanation:

The movement of the electron within the solenoid in a circle is produced by equaling the magnetic force and the centripetal force, as follows:

$F_{B} = F_{c}$

$e*v \mu_{0}*n*I = \frac{m*v^{2}}{r}$

$I = \frac{m*v}{e* \mu_{0}*n*r}$

Where:

I: is the current

m: is the electron's mass = 9.1x10⁺³¹ kg

v: is the electron's speed = 3.0x10⁵ m/s

μ₀: is the permeability magnetic = 4πx10⁻⁷ T.m/A

n: is the number of turns per unit length = 35/cm

r: is the radius of the circle = 3.0 cm

e: is the electron's charge = 1.6x10⁻¹⁹ C  

$I = \frac{m*v}{e*\mu_{0}*n*r} = \frac{9.1 \cdot 10^{-31} kg*3.0 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*4\pi \cdot 10^{-7} T.m/A*3500/m*0.03 m} = 0.0129 A$  

Therefore, the current of the solenoid is 0.0129 A.

I hope it helps you!