Question

The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the wire represent the direction of current). For the following wires, which all also carry current , indicate the magnitude (in mT) and direction of the magnetic field at the center (red point) of each configuration.

Answer 1

Complete Question

 The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is  $-\r k$

Explanation:

      From the question we are told that

                 The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

              $B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

So

                $B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is  (i.e the magnetic field generated)

                         $\frac{\mu_o I}{2R} = 1mT$

 This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

                   $B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and  for the larger semi-circular loop  is

                 $B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

    So the net magnetic field would be

                   $B_{net} = B_1 - B_2$

                        $= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

                        $=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

                        $=\frac{\mu_o I}{8R}$

                        $= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall  $\frac{\mu_o I}{2R} = 1mT$

    So  

             $B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$