A 0.780 m long section of cable carrying current to a car starter motor makes an angle of 60.0° with the Earth's 5.50 ✕ 10−5 T field. What is the current (in A) when the wire experiences a force of 7.00 ✕ 10−3 N?

The current is 188.17Amperes

Explanation:

The formula for calculating the force carrying current on a conductor placed in a magnetic field is expressed as shown;

F = BILsin(theta) where;

F is the force on the conductor = 7.00 ✕ 10^−3 N

B is the earth's magnetic field = 5.50 ✕ 10^−5 Tesla

I is the current = ?

L is the length of the conductor = 0.78m

theta is the angle that the conductor make with the earth's magnetic field = 60.0°

Making the current I the subject of the formula from the equation gives;

I = F/BLsin(theta)

Substituting the given data's in the formula to get current in the wire;

I = 7.00 ✕ 10^−3/(5.50 ✕ 10^−5 × 0.78 × sin60°)

I = 7.00×10^−3/3.72×10^-5

I = 188.17Amperes

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

d eras and periods