A long solenoid has a radius of 4.0 cm and has 800 turns/m.If the current in the solenoid is increasing at the rate of 3.0 A/s, what is the magnitude of the induced electric field at a point 2.2 cm from the axis of the solenoid?

Question

Quizplus · Accepted Answer

The induced electric field inside a solenoid is given by E = (μ₀ * n * r * (dI/dt)) / 2, where μ₀ is the permeability of free space, n is the number of turns per unit length, r is the distance from the axis, and dI/dt is the rate of change of current. Substituting the given values, E = (4π * 10^-7 T*m/A * 800 turns/m * 0.022 m * 3 A/s) / 2 = 3.3 * 10^-5 V/m.

A Long Solenoid Has a Radius of 4