Question

An electron orbits in a 0.5 mm radius circle path, with the magnetic moment of the orbiting electron being 4.2x10^-19 A m².

a) What is the magnitude of the electron current for the system?

b.) How long does it take the electron to make one orbit?

Answer 1

The magnetic moment of a current loop is given by,

Here, are magnetic moment, current in the loop and area of the loop.

Now here the radius of the loop is, 0.5mm=0.5x10^-3 m

So the magnetic moment is,

This is the electron current in the loop.

Now we know that the charge of the electron is

And the current is,

Here T is the time period.

So, it takes only 0.19 ms to make one orbit.

Answer 2

a) The magnetic moment of an electron orbiting in a circle is given by:

μ = IA

where μ is the magnetic moment, I is the current, and A is the area of the circle. Rearranging this equation to solve for I, we get:

I = μ / A

Plugging in the given values, we get:

I = (4.2 x 10^-19 A m^2) / (π x (0.5 x 10^-3 m)^2) I = 1.7 x 10^-11 A

Therefore, the magnitude of the electron current is 1.7 x 10^-11 A.

b) The time it takes for the electron to make one orbit is given by:

T = (2πr) / v

where T is the period, r is the radius of the circle, and v is the velocity of the electron. The velocity of the electron can be calculated from the centripetal force acting on it:

F = mv^2 / r qvB = mv^2 / r v = qBr / m

where q is the charge of the electron, B is the magnetic field strength, and m is the mass of the electron. Substituting this value for v into the equation for the period, we get:

T = (2πr) / (qBr / m) T = (2πm) / (qB)

Plugging in the given values, we get:

T = (2π x 9.11 x 10^-31 kg) / (1.602 x 10^-19 C x B) T = 3.57 x 10^-8 / B

Therefore, the time it takes for the electron to make one orbit depends on the strength of the magnetic field B. Without knowing the value of B, we cannot calculate the exact time.