# Magnetic field torque

## Homework Statement

a coil 2 cm in diameter has 300 turns. what is the maximum torque on this coil when it carries a current of 10mA and is placed ina constant magnetic field of 0.05 Teslas?

## Homework Equations

magnetic force on a current F = IL X B where X indicates cross product, I is current, L is length of coil, B is magnetic field

torque tau = rFsin(theta) where r is radius, F is magnetic force on current

## The Attempt at a Solution

basically subbed in IL X B for F in rFsin(theta)
and since r is radius, it is two times the diameter, 2*0.02 meters

so torque tau = (2*0.02)[(10*10^-3)(300)(0.05)sin(90)]sin(90) = 0.06 N/m

and since we won't max torque, force must be applied perpendicular, thus theta = 90 degrees.

i am unsure of the length L though, the problem states diameter is 2cm and contains 300 turns, how much is 300 turns in meters?

is my approach correct, though?

cheers

Dick
Homework Helper
There's really quite a lot of errors in there. To start with radius isn't 2*diameter. You're also using inappropriate formulas all over the place. What you want to find the force is the magnetic moment of the coil cross the applied field. What's a formula for the magnetic moment of a coil? You certainly can't convert 300 turns to meters, come on. You can use it to find the total current circulating. Start all over again with a clear head, ok?

i agree, it did seem a little too straightforward with my initial approach

magnetic dipole moment, mu = NIA where N is number of turns in coil, I is current, A is area( is it area of wire it self(a circle; A = pi(r^2) ?) --> I = 10*10^-3 ampere, N = 300, A = pi(r^2) = pi((2*0.02)^2) = 0.005 m^2

and torque, tau = mu X B where B is magnetic field = 0.05Teslas

once i determine area A, and then mu, will i then find torque?

so tau = mu X B = mu(B)sin(theta) = NIABsin(theta) = ((300)(10*10^-3)(0.005)(0.05))sin(theta) where theta = 90 degrees
---> ((300)(10*10^-3)(0.005)(0.05))sin(90) = 7.5*10^-4 N/m

correct now? not too sure about area, A though.

Dick
Homework Helper
You are making progress. Sure, area=pi*r^2. But you still have that weird radius=2*diameter thing going on. Can you do something about that? It's not right, is it? Yes, I think the question means you to take the coil to be a circle. Otherwise, they wouldn't give a 'diameter' right?

oh oh oh, i was getting confused with diameter and radius, diameter = 2 * radius, not the other other way round, so radius r = diameter/2 = 0.02/2 = 0.01 m

in that case

tau = mu X B = mu(B)sin(theta) = NIABsin(theta) = ((300)(10*10^-3)(3.14*10^-4)(0.05))sin(90) = 4.71*10^-5 newton meters

Dick