Question

A single-turn rectangular wire loop measures 6.00 cm wide by 10.0 cm long. The loop carries a current of 3.00 A. The loop is in a uniform magnetic field with B = 4.00 times 10^-3 T. Taking torques about an axis, parallel to either side of the rectangular loop, that maximizes the torque, what is the magnitude of the torque exerted by the field on the loop if the direction of the magnetic field is described as the following? a) parallel to the short sides of the loop N middot m b) parallel to the long sides of the loop N middot m c) perpendicular to the plane of the loop N middot m

Answer 1

Answer:

Explanation:

Single turn implies that N=1

The rectangle is 6cm (0.06m) wide and 10 cm (0.1m) long.

Current in loop is 3A

Magnetic Field B= 4×10^-3T

To take the maximum torque then, sinθ=1

Then

τ=NiABsinθ

Where N is number of turns, and in this case N is 1

i is current in coil and it this case it is 3Amps.

A is area of of coil

And in this case it is a rectangle

Area of rectangle is Lenght × breadth

A= 0.06×0.1= 0.006m^2

And B is magnetic field given as B=4×10^-3T

And θ is angle between the field and the normal to the coil.

a. Torque parallel to the short side of the loop,

The length of the short side is 6cm=0.06m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=1×3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm

b. Torque parallel to the long side of the loop,

The length of the short side is 10cm=0.1m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm.

c. Torque perpendicular to the plane. When the field is perpendicular to the loop the torque is zero.

If the torque is perpendicular to the plane, we said theta is the angle between the normal and the magnetic field

Then if the torque is perpendicular to the normal, then the angle between the torque and the normal is 0, the sinθ = 0

τ=NiABsinθ

Since sinθ =0

Then,

τ=0Nm