Question

A straight wire carries a current of 40 A in a uniform magnetic field (magnitude = 80 mT). If the force per unit length on this wire is 2.0 N/m, determine the angle between the wire and the magnetic field.

Answer 1

Answer:

38.68 °C

Explanation:

Using,

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

F/L = BIsinФ...................... Equation 2

Where F/L = force per unit length, B = Magnetic Field, I = current, Ф = angle between the wire and the magnetic Field

make Ф the subject of the equation

Ф = sin⁻¹[(F/L)/BI]................... Equation 3

Given: F/L = 2 N/m, B = 80 mT = 0.08 T, I = 40 A.

Substitute into equation 3

Ф = sin⁻¹(2/(0.08×40))

Ф = sin⁻¹(0.625)

Ф = 38.68 °C

Hence the angle between the wire and the magnetic Field = 38.68 °C

Answer 2

Answer:

38.7°

Explanation:

The force on a current carrying wire is given as

F = (B)(I)(L) sin θ

where

B = magnetic field = 80 mT = 0.08 T

I = current in the wire = 40 A

(F/L) = 2.0 N/m

θ = angle between the magnetic field and the direction of current flow (basically angle between the wire and the magnetic field) = ?

(F/L) = (B) (I) sin θ

2 = 0.08 × 40 × sin θ

sin θ = (2/3.2) = 0.625

θ = sin⁻¹ (0.625)

θ = 38.7°

Hope this Helps!!!