A square loop of wire is held in a uniform 0.35 T magnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of 2.8 cm/s. What emf is induced in the loop when the length is 8.3 cm

-1.627×10⁻³ V.

Explanation:

Applying,

E = -dФ/dt Equation 1

Where E = induced emf, dФ = change in magnetic flux, dt = change in time.

But,

Ф = BA

Where, B = magnetic Field, A = Area of the loop

Since the loop is a square,

A = l²

Where l = length of the loop

Therefore,

Ф = Bl² Equation 2

Substitute equation 2 into equation 1

E = -d(Bl²)/dt

E = -2Bldl/dt Equation 3

Given: B = 0.35 T, l = 8.3 cm = 0.083 m, dl/dt = 2.8 cm/s = 0.028 m/s.

Substitute into equation 3

E = -2(0.35×0.083×0.028)

E = -1.627×10⁻³ V.

Hence the induced emf =  -1.627×10⁻³ V

Note: The negative sign shows that the induced emf opposes the motion that produces it.

The induced emf in the loop is 1.627 mV

Explanation:

Given;

magnetic field strength, B = 0.35 T

the length of each side of the square is decreasing at a constant rate, dl/dt = 2.8 x 10⁻² m/s

Induced emf is given as;

where;

Ф is magnetic flux through the loop, given as;

Ф = BA

A is area, area of square = L²

Ф = BL²

Substitute in the given values;

Therefore, the induced emf in the loop is 1.627 mV

the answer is a 3.8cm

explanation: