Physics 2
posted by Megan on .
A wire has a length of 4.52 x 102 m and is used to make a circular coil of one turn. There is a current of 7.94 A in the wire. In the presence of a 8.04T magnetic field, what is the maximum torque that this coil can experience?

A 45 turn coil of radius 5.1 cm rotates in a uniform magnetic field having a magnitude of 0.49 T. If the coil carries a current of 40 mA, find the magnitude of the maximum torque exerted on the coil.

Torque = (the number of turns)(current)(Area)((magnetic field)(sin(angle)))
Area = π × r^2
the angle is the angle between the normal to the plane of the coil and the magnetic field.
Torque = (45)(40)(π(5.1)^2)(0.49 sin(90))
Maximum torque occurs when the magnetic force is perpendicular to the normal (which is sin(90)= 1
= (45)(40)(π(5.1)^2)(0.49)
= 72070.71158 N*m 
@ Megan

Torque = (# of turns)(current)(Area)((magnetic field)(sin(angle)))
Area = (circumference)^2 / (4π)
circumference is distance around the edge of a circle.
Torque = (1)(7.94)(((4.52 x 10^2)^2)/(4π))(8.04 sin(90))
=(7.94)(((4.52 x 10^2)^2)/(4π))(8.04)
=(7.94)((.0452)^2/(4π))(8.04)
= .0103787 N*m
WHEN YOU CALCULATE AREA ON YOUR CALCULATOR MAKE SURE TO PUT BRACKETS AROUND THE 4π.
****THIS > (.0452)^2/(4π)
NOT THIS > (.0452)^2/4π
IF YOU DON'T ADD BRACKETS THEN IT WILL NOT GIVE YOU THE CORRECT AREA.