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? A) 3.3 × 10−⁵ V/m B) 3.6 × 10−⁵ V/m C) 3.9 × 10−⁵ V/m D) 4.2 × 10−⁵ V/m E) 6.0 × 10−⁵ V/m

Question

Quizplus · Accepted Answer

The induced electric field at a point outside the solenoid can be calculated using the formula:
E = μ₀ * n * ΔI * r / 2
where μ₀ is the permeability of free space, n is the number of turns per unit length, ΔI is the change in current, and r is the distance from the axis of the solenoid.

Substituting the given values, we get:
E = (4π × 10⁻⁷ T•m/A) * (800 turns/m) * (3.0 A/s) * (0.022 m) / 2
E ≈ 3.3 × 10⁻⁹ V/m
Therefore, the correct answer is choice A.

A Long Solenoid Has a Radius of 4