Question

For a particular scientific experiment, it is important to be completely isolated from any magnetic field, including the earth's field. The earth's field is approximately 50 μT, but at any particular location it may be a bit more or less than this. A 1.00-m-diameter current loop with 200 turns of wire is adjusted to carry a current of 0.211 A ; at this current, the coil's field at the center is exactly equal to the earth's field in magnitude but opposite in direction, so that the total field at the center of the coil is zero.
What is the strength of the earth's magnetic field at this location?
Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

The magnetic field at the center of the coil is   $B =53 \mu T$

Explanation:

   From the question we are told that

         The magnetic field of earth is $B_e = 50 \mu T = 50 *0^{-6} \ T$

         The diameter of the wire is  $d = 1.0\ m$

        The radius is $r = \frac{1}{2} = 0.5 \ m$

         The number of turn is  $N = 200 \ turns$

         The current it is carrying is  $I = 0.211 \ A$

The earth magnetic field strength at the center of the coil is mathematically represented as

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

Where $\mu_o$ is the permeability of free space

           $\mu_o = 4\pi * 10^{-7} N/A^2$

So  

         $B = 200* \frac{ 4\pi * 10^{-7} * 0.211 }{2 * 0.5}$

       $B = 5.304*10^{-5}$

        $B =53 \mu T$