A wire has a length of 4.52 x 10-2 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 8.04-T magnetic field, what is the maximum torque that this coil can experience?

Answer 1

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.

Answer 2

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

Answer 3

@ Megan
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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.