Chapter 4: Q38E (page 984)

A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the centre of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Short Answer

Expert verified

(a) The magnitude of the induced electric field at the point near the centre of the solenoid and 0.500cm from the axis is $1.20 \times 10^{- 4} V / m$

(b) The magnitude of the induced electric field at the point near the centre of the solenoid and from the axis is $2.04 \times 10^{- 4} V / m$

Step by step solution

Electric field at a distance r

To find the induced electric field E at a distance r from the centre of the solenoid. The magnetic field at the centre of the solenoid is:

$B = μ_{0} n l$

From Faraday’s law we get:

$\oint \vec{E} . \vec{d l} = E (2 π r)$

To calculate the magnitude of E we take the absolute value therefore:

$E (2 π r) = π r^{2} |\frac{d B}{d t}|$

We can also write it as:

$\frac{d B}{d t} = μ_{0} n \frac{d l}{d t}$

Calculating the values of the electric field at the given distances from the solenoid

At 0.500cm

Now to calculate the value we just substitute the value in the above equation we get:

$E = \frac{1}{2} (0.00500) (4 π 10^{- 7}) (900) (36.0)$

At distance

$E = \frac{1}{2} (0.0100) (4 π 10^{- 7}) (900) (36.0) = 1.02 \times 10^{- 4} V / m$

Therefore the electric field at is $1.02 \times 10^{- 4} V / m$

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Short Answer

Step by step solution

Electric field at a distance r

Calculating the values of the electric field at the given distances from the solenoid

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