Question

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

Answer 1

Answer:

E = 3.3 x 10^(-5) V/m

Explanation:

The time-dependent magnetic field produces an electric field. Due to symmetry of the problem, the magnitude of the electric field depends only on the distance from the axis of the solenoid. It's 2.2cm from the solenoid axis. We introduce an imaginary circular loop with the radius of 2.2 cm and the center at the axis of the solenoid. Since the magnitude of the magnetic field is constant at all points of the loop, then we can find the corresponding induced emf:

ε = EL

Now, the rate of change of the magnetic field is related to the rate of change of current in the solenoid as;

dB/dt = μo•n(dI/dt)

Where;

μ0 is the permeability of free space and has a value of 4π × 10^(-7) W/A².m

n is the number of turns per unit length

(dI/dt) is change of current in solenoid.

Thus,

dB/dt = 4π × 10^(-7)•800•3

dB/dt = 0.003 T/s

Now, from earlier we saw that,

ε = EL

Here, L is the circumference of the loop, which is L = 2πr and r = 2.2cm = 0.022m.

E is the Electric field

Thus, ε = 2πrE = 0.044πE - - - - (eq1)

The induced emf can be also found from the Faraday’s law as the rate of change of the magnetic flux through the circular loop motion as;

ε = -dΦ/dt

Now, the magnetic flux is the product of the magnetic field and the area of the circular loop speed.

Thus, Φ = BA

Where A is area = πr²

Thus, ε = -dΦ/dt = - AdB/dt

ε = -πr²(dB/dt) - - - - - (eq2)

So equating equation 1 and 2,we have;

0.044πE = -πr²(dB/dt)

π will cancel out to give;

0.044E = -r²(dB/dt)

Plugging in the relevant values ;

0.044E = -(0.022²)(0.003)

E = -0.000001452/0.044

E = -3.3 x 10^(-5) V/m

The negative sign is the Lenz law. It determines the direction of the induced electric field. The absolute value of the magnitude of induced electric field is; E = 3.3 x 10^(-5) V/m

Answer 2

Answer:

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.