Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnetic field is [tex]B_{net} = \frac{1}{4} * mT[/tex]
And the direction is [tex]-\r k[/tex]
Explanation:
From the question we are told that
The magnetic field at the center is [tex]B = 1mT[/tex]
Generally magnetic field is mathematically represented as
[tex]B = \frac{\mu_o I}{2R}[/tex]
We are told that it is equal to 1mT
So
[tex]B = \frac{\mu_o I}{2R} = 1mT[/tex]
From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)
[tex]\frac{\mu_o I}{2R} = 1mT[/tex]
This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)
[tex]B_1 = \frac{1}{2} \frac{\mu_o I}{2R}[/tex]
and for the larger semi-circular loop is
[tex]B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
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
[tex]B_{net} = B_1 - B_2[/tex]
[tex]= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
[tex]=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}[/tex]
[tex]=\frac{\mu_o I}{8R}[/tex]
[tex]= \frac{1}{4} \frac{\mu_o I}{2R}[/tex]
Recall [tex]\frac{\mu_o I}{2R} = 1mT[/tex]
So
[tex]B_{net} = \frac{1}{4} * mT[/tex]
Using the Right-hand rule we see that the direction is into the page which is [tex]-k[/tex]
