Question

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70 below the horizontal north direction. Find the current in the loop.

Answer 1

Answer:

The current is  $I = 6.68 \ A$

Explanation:

From the question we are told that  

     The radius of the loop is  $r = 6 \ cm = 0.06 \ m$

     The  earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

      The  number of turns is  $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

        $B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         $B = B_e$

=>     $B_e = \frac{\mu_o * N * I }{ 2 * r}$

     Where  $\mu$ is the permeability of free space with value  $\mu _o = 4\pi * 10^{-7} N/A^2$

       $0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=>     $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

       $I = 6.68 \ A$