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

Answer:

The maximum torque is  $\tau_{max} = 1.139 *10^{-7} \ N \cdot m$

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

        $\mu = \frac{I l^2}{4 \pi n }$

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

    substituting values

       $\mu = \frac{ 5.77 *10^{-3} * 0.26^2}{4 * 3.142* 1 }$

     $\mu = 3.10 4* 10^{-5} A m^2$

Now  

      $\tau_{max} = 3.104 *10^{-5} * 3.67 *10^{-3}$  

      $\tau_{max} = 1.139 *10^{-7} \ N \cdot m$