Question

A long, thin solenoid has 700 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s . a. What is the magnitude of the induced electric field at a point near the center of the solenoid? b. What is the magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid?

Answer 1

To solve this problem it is necessary to apply the concepts related to the magnetic field in a solenoid.

By definition we know that the magnetic field is,

$B = \mu_0 n I$

$\frac{dB}{dt} = \frac{d}{dt} \mu_0 NI$

$\frac{dB}{dt} = \mu_0 n\frac{dI}{dt}$

At the same tome we know that the induced voltage is defined as

$\epsilon = \frac{\Phi}{dt}$

$\epsilon = A \frac{dB}{dt}$

Replacing

$\epsilon = A \mu_0 n\frac{dI}{dt}$

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}$

PART A) Substituting with our values we have that

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0)(36)}{2}\\\epsilon = 0V/m$

Therefore there is not induced electric field at the center of solenoid.

PART B) Replacing the radius for 0.5cm

$\epsilon = \frac{\mu nr}{2} \frac{dI}{dt}\\\epsilon = \frac{(4\pi*10^{-7})700(0.5*10^{-2})(36)}{2}\\\epsilon = 7.9168*10^{-5}V/m$

Therefore the magnitude of the induced electric field at a point 0.5cm is $7.9168*10^{-5}V/m$

Answer 2

Answer:

a) The magnitude of the induced electric field at a point near the center of the solenoid is zero (0)

b) The magnitude of the induced electric field at a point 0.500 cm is 7.92x10⁻⁵V/m

Explanation:

Please look at the solution in the attached Word file

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