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

The magnitude of the induced electric field is 3.32 x 10⁻⁵ V/m.

The given parameters;

The magnitude of the magnetic field strength is calculated as follows;

B = μ₀nI

The magnitude of the induced electric field is calculated as follows

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Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

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