Question

A solenoid with 33 turns per centimeter carries a current I. An electron moves within the solenoid in a circle that has a radius of 2.1 cm and is perpendicular to the axis of the solenoid. If the speed of the electron is 3.6 ✕ 105 m/s, what is I (in A)?

Answer 1

To develop this problem it is necessary to match the concepts related to electromagnetic force and the centripetal Force.

By definition we have that the centripetal Force is equivalent to

$F_c = \frac{mv^2}{R}$

Where,

m = Mass (of a electron)

v = Velocity

R = Radius

At the same time we have that magnetic force is equal to

$F_e = qvB$

Where,

q = Charge

V = Velocity

B = Magnetic Field

Equating both we have,

$F_c = F_e$

$\frac{mv^2}{R} = qvB$

Re-arrange to find B,

$B = \frac{mv}{qR}$

Replacing with our values we have,

$B = \frac{(9.1*10^{-31})(3.6*10^5)}{(1.6*10^{-19})(2.1*10^{-2})}$

$B = 0.0000975T$

Now for Faraday's law the Magnetic field in a solenoid is defined as,

$B = \mu_0 NI$

Re-arrange to find I

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

Where,

B = Magnetic Field

$\mu =$ Permeability constant

N = Number of loops per meter

Replacing with our values

$I = \frac{(0.0000975)}{(4\pi*10^{-7})(3300)}$

$I = 0.235115A$

Therefore the Current is 0.235115A