Question

A wire of length 26.0 cm carrying a current of 5.77 mA is to be formed into a circular coil and placed in a uniform magnetic field B of magnitude 3.67 mT. If the torque on the coil from the field is maximized, what is the magnitude of that maximum torque

Answer 1

The maximum torque is
`\tau_(max) = 1.139 *10^(-7) \ N \cdot m`

Step-by-step explanation:

From the question we are told that

The length of the wire is
`l = 26.0 \ cm = 0.26 \ m`

The current flowing through the wire is
`I = 5.77mA = 5.77 *10^(-3) \ A`

The magnetic field is
`B = 3.67 \ mT = 3.67 *10^(-3 ) T`

The maximum torque is mathematically evaluated as

`\tau_(max) = \mu B`

Where
`\mu` is the magnetic dipole moment which is mathematically represented as

`\mu = (I l^2)/(4 \pi n )`

Where
`n` is the number of turns which from the question is 1

substituting values

`\mu = ( 5.77 *10^(-3) * 0.26^2)/(4 * 3.142* 1 )`

`\mu = 3.10 4* 10^(-5) A m^2`

Now

`\tau_(max) = 3.104 *10^(-5) * 3.67 *10^(-3)`

`\tau_(max) = 1.139 *10^(-7) \ N \cdot m`