12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4  = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.