A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)

Explanation:

The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .

a )

Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .

b )

magnetic field created at the magnetic needle B = 10⁻⁷ x  2I / d where I is current and d is distance .

B = 10⁻⁷ x  2 x 26.3  / .27

= 194.81 x 10⁻⁷ T

angle of deflection of solenoid = 22.9°

Tan 22.9 = B /H

.422 = 194.81 x 10⁻⁷ / H

H = 461.63 x 10⁻⁷ T

= .46 x 10⁻⁴ T .

the answer is 46 mls because when you add the speed per second it increses the speed

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