Question

A long, thin solenoid has 950 turns per meter and radius 3.00 cm. The current in the solenoid is increasing at a uniform rate of 63.0 A>s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Answer 1

Answer:

Part a)

$E = 0.188 \times 10^{-3} N/CPart b)[tex]E = 0.376 \times 10^{-3}$N/C

Explanation:

As we know that the magnetic field near the center of the solenoid is given as

$B = \mu_0 ni$

Also we know by  equation of Faraday's law

EMF induced in the closed loop will be equal to rate of change in magnetic flux

so we have

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

so we have

$\int E. dL = \pi r^2 (\mu_0 n \frac{di}{dt})$

$E. (2\pi r) = \pi r^2 (\mu_0 n \frac{di}{dt})$

$E = \frac{\mu_0 n r}{2} \frac{di}{dt}$

Part a)

At r = 0.500 cm

we have

$E = \frac{4\pi \times 10^{-7} (950) (0.500 \times 10^{-2})}{2}(63)$

$E = 0.188 \times 10^{-3} N/C$

Part b)

At r = 1.00 cm

we have

$E = \frac{4\pi \times 10^{-7} (950) (1.00 \times 10^{-2})}{2}(63)$

$E = 0.376 \times 10^{-3}$N/C