A square frame with a side of 3 m was placed in a uniform magnetic field with an induction of 1 T perpendicular

A square frame with a side of 3 m was placed in a uniform magnetic field with an induction of 1 T perpendicular to the lines of induction, 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: 2. What charge passed along the circuit in this case? Frame resistance 1 ohm.

a = 3 m.

B = 1 T.

R = 1 ohm.

c: b = 1: 2.

∠α = 0 °.

Q -?

I = EMF / R.

I = Q / t.

EMF = ΔF / t.

Q / t = ΔФ / t * R.

Q = ΔФ / R.

ΔФ = В * ΔS * cos∠α.

S1 = a ^ 2.

S1 = (3 m) ^ 2 = 9 m ^ 2.

Find the perimeter of the square frame, the length of the wire: P = 4 * a.

P = 4 * 3 m = 12 m.

Let us express the area of the rectangular frame: S2 = c * b.

2 * c = b.

P = 2 * (c + b) = 2 * (c + 2 * c) = 6 * c.

c = 2 m, b = 4 m.

S2 = 2m * 4m = 8m ^ 2.

ΔS = S1 – S2.

ΔS = 9 m ^ 2 – 8 m ^ 2 = 1 m ^ 2.

Q = 1 T * 1 m ^ 2/1 Ohm = 1 Cl.

Answer: a charge of Q = 1 C will pass through the conductor.