# Use your measurements from part e to calculate the strength of the earth’s magnetic field

Magnetic field = 5.66 x 10^ -6

The given question is to determine earth's magnetic field.

When a current I amp. flows through the wire a magnetic field B
is developed.

This magnetic field is perpendicular to horizontal component of
earth's magnetic field H.

If a compass needle is placed in the external magnetic field B ,
it experiences a resultant force and lies at an angle
with the horizontal component of earth's magnetic field.

According to Tangent law B = Htan

And H = B / tan.

Calculating the external magnetic
field B .

We can use Biot Savartt law to calculate magnetic field due to
the current.

According to Biot Savartt law , the magnetic field B =
0 I / 2π R.

Where
0 = permeability of vacuum.

0
= 1.256 * 10-6

I = current , the given Value = 1.98 A.

And R = distance between the wire and the point where the field
is calculated.

R = 7.0 cm., Given in the problem.

R = 0.07 m.

Substituting the given values for calculating magnetic
field,

B = 1.256 * 10-6 x 1.98 / 2* 3.14 * 0.07

B = 5.6569 * 10-6

Calculating earth's magnetic
field.

Applying Tangent law B = H tan

The angle at which the compass needle makes with horizontal
component of earth's magnetic field , the given Value =
450

Hence the earth's magnetic field H = B/ tan

H = 5.6569 * 10-6 / tan45 .

The value of tan45 = 1.

Hence the earth's magnetic field

H= 5.6569 * 10-6 T.