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°

(a) 2.42 N

(b) 2.8 N

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

(a) 2.425 N

(b) 2.8 N

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

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .