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.
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
[tex]\frac{dB}{dt} = \mu_0 n\frac{di}{dt}[/tex]
[tex]\frac{dB}{dt} = (4\pi \times 10^{-7}) \times (800) \times 3 = 0.00302 \ T/s[/tex]
The magnitude of the induced electric field is calculated as follows
[tex]E = \frac{r}{2} (\frac{dB}{dt} )\\\\E = \frac{0.022}{2} \times (0.00302)\\\\E = 3.32 \times 10^{-5} \ V/m[/tex]
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