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 0.203 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.

Answer 1

Answer:

51.019 μT

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

i = Current in wire = 0.203 A

N = Number of turns = 200

d = Diameter = 1 m

Magnetic field is given by

$B=\dfrac{N\mu_0i}{d}\\\Rightarrow B=\dfrac{200\times 4\pi \times 10^{-7}\times 0.203}{1}\\\Rightarrow B=0.000051019\ T$

The strength of the magnetic field at this location would be 51.019 μT

Answer 2

The magnetic field at the given point is 51.019 μT.

Magnetic field:

It is given that the current in the coil is I = 0.203A

the diameter of the loop carrying the current is d = 2r = 1m

and the number of turns is N = 200.

The magnetic field to a current-carrying loop having N number of turns and carrying a current I with the radius of the loop being r is given by:

$B=\frac{\mu_oNI}{2r} \\\\B=\frac{4\pi\times10^{-7}\times200\times0.203}{1}\\\\B=51.019\;\mu T$

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