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

The magnitude of the induced emf near the center of the solenoid is 0 V/m.

The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is 8 x 10⁻⁵ V/m.

The given parameters;

• number of turns of the solenoid, N = 700 turns/m
• radius of the wire, r = 2.5 cm
• current in the solenoid, I = 36 A/s

The magnitude of the induced emf near the center of the solenoid is calculated as follows;

$B = \mu_0 nI\\\\\frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \\\\\frac{dB}{dt} = (4\pi \times 10^{-7} \times 700 \times 36) = 0.032 \ T/s$

$E = \frac{r}{2} (\frac{dB}{dt} )\\\\\ E = \frac{0}{2} (0.032)\\\\ E = 0 \ V/m$

The magnitude of the induced electric field at a point 0.500 cm from the axis of the solenoid is calculated as follows;

$E = \frac{r}{2} (\frac{dB}{dr} )\\\\ E= \frac{0.5\times 10^{-2} }{2}( 0.032)\\\\E = 8\times 10^{-5} \ V/m$

Learn more here:https://brainly.com/question/15399686