A particle of charge q moves in a circle of radius a with speed v. Treat the circular path as a current loop with a constant current equal to the ratio of the particles charge magnitude to its period of motion. Find the maximum possible magnitude of the torque produced on the loop by a uniform magnetic field of magnitude B. (Use any variable or symbol stated above as necessary. Assume all quantities are in SI units.)

Question

Luca Marquez

Physics

8 April, 04:22

A particle of charge q moves in a circle of radius a with speed v. Treat the circular path as a current loop with a constant current equal to the ratio of the particles charge magnitude to its period of motion. Find the maximum possible magnitude of the torque produced on the loop by a uniform magnetic field of magnitude B. (Use any variable or symbol stated above as necessary. Assume all quantities are in SI units.)

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Answers (1)

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Diya · Answer 1

The magnitude is 6.87 * 10^-26N. m

Explanation:

There are missing details in this question.

To solve this question, I'll assume the values of radius, speed and the magnitude of magnetic speed.

Let r = radius = 5.3 * 10^-11m

Let v = speed = 2.2 * 10^6 m/s

Magnitude = 7.0mT

First, we'll calculate the period of the motion ...

This is calculated using 2πr/v

T = 2 * π * 5.3 * 10^-11/2.2 * 10^6

T = 1.513676460365E-16

I = 1.514 * 10^-16

Then we calculate the current.

Current, I = e/T where T = 1.69 * 10^-19

So I = 1.69 * 10^-19/1.514 * 10^-16

I = 0.001116248348745

I = 0.001112 A

Next we calculate area

Area = πr²

Area = π (5.3 * 10^-11) ²

Area = 8.824733763933E-21

Area = 8.825E-21 m²

Lastly, the maximum magnitude produced by the torque is calculated as:.

i * Area * Magnitude

0.001112 * 8.825 * 10^-21 * 7 * 10^-3

6.86938E-26.

Hence, the magnitude is 6.87 * 10^-26N. m