A straight wire that is 0.56 m long is carrying a current of 2.6 A. It is placed in a uniform magnetic field, where it experiences a force of 0.24 N. The wire makes an angle of 900 with the magnetic field. What is the magnitude of the magnetic field

Explanation:

The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;

B is the magnetic field (in Tesla)

I is the current (in amperes)

L is the length of the wire (in meters)

∝ is the angle that the conductor makes with the magnetic field.

Given parameters

L = 0.56 m

I = 2.6A

F = 0.24N

∝  = 90°

Required

magnitude of the magnetic field (B)

Substituting the given values into the formula given above we will have;

F = BILsin∝

0.24 = B * 2.6 * 0.56 sin90°

0.24 =  B * 2.6 * 0.56 (1)

0.24 = 1.456B

1.456B = 0.24

Dividing both sides by 1.456 will give;

1.456B/1.456 = 0.24/1.456

B ≈ 0.165Tesla

Hence the magnitude of the magnetic field is approximately 0.165Tesla

think its the third option

a. are