Question

5 A. The current flowing in a solenoid, of 400 turns, 20 cm length & 4 cm diameter, changes with time according to the graph show to right. Derive an expression for the strength of the induced electric field inside the solenoid. 0 0.0 01 02 03 0.4 Sketch the corresponding graph showing how the induced electric field vary with time.

Answer 1

The magnetic flux through the solenoid, $\phi$ = B * A
Where B is the magnetic field and A is the area.
B = ($\mu$o * N * I) / L
Where N is the number of turns, I is the current and L is the length of the solenoid
A = $\pi$ * R²
Where R is the radius of the solenoid.
_$\phi$ = ($\mu$o * N * I) / L * ($\pi$ * R²)
d_$\phi$/dt = [($\mu$o * N * $\pi$ R²) / L] * dI/dt
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Using Faraday's law, the induced electric field can be written as,
E * 2$\pi$r = - d_$\phi$/dt
Where r is the distance from the center of the solenoid to the point where we need to know the electric field.
E = - [($\mu$o * N * R²) / (L * r)] * dI/dt
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E = k * - dI / dt, where k = [($\mu$o * N * R²) / (L * r)]
From the graph, dI / dt = 5 / 0.1 from 0 to 0.1 s
E = - 50 k

dI dt = 0 from 0.1 to 0.2 s, E = 0

dI/dt = 5 / (0.2 - 0.4) = - 25 from 0.2 s to 0.4 s,
E = + 25k

5 -t (s) 0 0.0 0.1 0.2 0.3 0.3 0.4 E 25k 0 t(s) - 50k