Question

A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.150 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A.
(a) Find the force on each side of the loop.
magnitude = µN.

(b) Find the magnitude of the torque acting on the loop.
= N · m

Answer 1

Answer:

a

The force acting on one of the  sides  the square loop is 169.6 N

b

The torque acting on the loop is ZERO

Explanation:

   From the question we are given that

            permeability of free space   $\mu_0$  $= 4\pi*10^{-7} T\cdot m/A$

         Number of turns of the solenoid per unit length , n = 30 turns per cm

                                                                                             = 3000 turns per m

          The current flowing through the solenoid  $I_1$ = 15A

          Length of the side of the loop, L = 2 cm

                                                                 =  2  cm [$\frac{1*10^{-2}m}{1cm}$]

                                                                  =  $2*10^{-2} m$

                 Angle ,  $\theta =90$°  

                 The current flowing through the wire $I_2 = 0.15A$    

Generally The formula for  Magnetic field inside of a solenoid is  

          $B = \mu_0nI_1$

Where

           $\mu_0$ is the permeability of  free space

            n is the number of turns of the solenoid per unit length

            $I_1$ is the current through the solenoid

Also the formula for force acting on one of the sides of a square loop is

            $F = BI_2Lsin \theta$

  Where  $I_2$ is the current through the wire ,

   L is the length  of the sides of the loop,and

    $\theta$ is the angle between B and $I_2$

Substituting the given values into the first formula

           $B = (4 \pi *10^{-7})(3000)(15)$

               $= 5.65*10^{-2} T$

To obtain the force acting on the loop by substituting the appropriate value into the second formula

                 $F = (5.65*10^{-2}(0.15)(2*10^{-2}(sin90)))$

                     $169.6 \mu N$

   The torque acting on the lop  is zero because the

          The forces on the loop are acting in the plane of the loop

          These force are perpendicular to the sides

          and their direction is away from the inner side of the loop

         the only effect the orientation of these forces can cause is stretching of the loop

       This way by which the forces are oriented cannot cause any rotation