A square of wire with a resistance of 5 ohms was placed in a uniform magnetic field with an induction

A square of wire with a resistance of 5 ohms was placed in a uniform magnetic field with an induction of 0.2 T perpendicular to the induction lines, then, without removing the wire from the field and without changing its orientation, it was deformed into a rectangle with a side ratio of 1: 3. In this case, a charge of 4 μC passed along the contour. What is the length (in cm) of the wire?

B = 0.2 T.

R = 5 ohms.

c: b = 1: 3.

∠α = 0 °.

Q = 4 μC = 4 * 10 ^ -6 C.

R – ?

I = EMF / R.

I = Q / t.

EMF = ΔF / t.

Q / t = ΔФ / t * R.

Q = ΔФ / R.

ΔФ = В * ΔS * cos∠α.

Q = B * ΔS * cos∠α / R.

Let us express the change in the area of the frame during deformation: ΔS = Q * R / B * cos∠α.

ΔS = 4 * 10 ^ -6 C * 5 Ohm / 0.2 T = 0.0001 m ^ 2.

Find the area and perimeter of the square: S1 = a ^ 2, P = 4 * a.

Let’s find the area and perimeter of the rectangle: S2 = c * b, P = 2 * (c + b) = 2 * (c + 3 * c) = 8 * c.

4 * a = 8 * c.

a = 2 * c.

b = 3 * c.

ΔS = S1 – S2 = a ^ 2 – c * b = (2 * c) ^ 2 – c * 3 * c = 4 * c ^ 2 – 3 * c ^ 2 = c ^ 2.

c = √ΔS.

c = √0.0001 m ^ 2 = 0.01 m.

b = 3 * 0.01 m = 0.03 m.

P = 2 * (0.01 m + 0.03 m) = 0.08 m = 8 cm.

Answer: the length of the wire is P = 8 cm.