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

The current of the solenoid is 0.0129 A.

Step-by-step 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 = (m*v^(2))/(r)`

`I = (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 = (m*v)/(e*\mu_(0)*n*r) = (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!