Question

Let's apply the torque equation to a circular coil of wire in a magnetic field. The coil has an average radius of 0.0435 m , has 40 turns, and lies in a horizontal plane. It carries a current of 7.25 A in a counterclockwise sense when viewed from above. The coil is in a uniform magnetic field directed toward the right, with magnitude 1.05 T . Find the magnetic moment and the torque on the coil. Which way does the coil tend to rotate

Answer 1

`M=1.72Am^2`

`T=1.810N-m`

Step-by-step explanation:

From the question we are told that:

Radius
`r=0.0435`

Number of turns
`N=40turns`

Current
`I= 7.25 A`

Magnetic Field
`B=1.05T`

Generally the equation for magnetic moment M is mathematically given by

`M=NIA`

Where

`A= Surface Area`

`A=\pi(r)^2`

`A=3.142*(0.0435)^2`

`A=5.95*10^(-3)`

Therefore the magnetic moment is given as

`M=NIA`

`M=40*7.25*5.95*10^(-3)`

`M=1.72Am^2`

Generally the equation for Torque on the coil T is mathematically given by

`T=M*B`

`T=1.72*1.05`

`T=1.810N-m`

Therefore torque on coil T is given a

`T=1.810N-m`