Question

A wire 2.80 m in length carries a current of 4.00 A in a region where a uniform magnetic field has a magnitude of 0.250 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a) 60.0°
(b) 90.0°

Answer 1

(a) 2.425 N

(b) 2.8 N

Step-by-step explanation:

(a)

The force on wire carrying current in a magnetic field is given as,

F = BILsinФ....................... Equation 1

Where F = magnetic force on the wire, B = Magnetic Field, I = Current, L = Length, Ф = angle between the magnetic Field and the current.

Given: B = 0.25 T, L = 2.8 m, I = 4 A, Ф = 60°

Substitute into equation 1

F = 0.25(4)(2.8)sin60°

F = 2.8×0.866

F = 2.425 N.

(b)

Similarly,

F = BILsinФ

Given: B = 0.25 T, L = 2.8 m, I = 4 A, Ф = 90°.

Substitute into the equation above

F = 0.25(2.8)(4)sin90°

F = 2.8(1)

F = 2.8 N

Answer 2

(a) 2.42 N

(b) 2.8 N

Step-by-step explanation:

Parameters given:

Current, I = 4A

Length of wire, L = 2.8m

Magnetic field strength, B = 0.25 T

Magnetic force is given as:

F = I*L*B*sinθ

(a) When θ = 60°:

F = 4 * 2.8 * 0.25 * sin60

F = 2.42N

(b) When θ = 90°:

F = 4 * 2.8 * 0.25 * sin90

F = 2.8N