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.

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!!!

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

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.