# A technician wraps wire around a tube of length 33 cm having a diameter of 8.1 cm. When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 590 turns of wire.

1 answer

###### Question:

A technician wraps wire around a tube of length 33 cm having a diameter of 8.1 cm. When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 590 turns of wire. (a) Find the self-inductance of this solenoid. (b) If the current in this solenoid increases at the rate of 3 A/s, what is the self-induced emf in the solenoid?

## Answers

Explanation:It is given that,Length of the tube, l = 33 cm = 0.33 mDiameter of the tube, d = 8.1 cm = 0.081 mNumber of turns, N = 590(a) We need to find the self- inductance of the solenoid. The formula for the self inductance is given by :[tex]L=\dfrac{\mu_o N^2A}{l}[/tex]A is the area of the solenoid[tex]L=\dfrac{4\pi \times 10^{-7}\times (590)^2\times \pi \times (0.0405)^2}{0.33}[/tex]L = 0.00683 HorL = 6.83 mHSo, the self inductance of the solenoid is 6.83 mH.(b) The current in this solenoid increases at the rate of 3 A/s. [tex]\dfrac{dI}{dt}=3\ A/s[/tex]EMF in the solenoid is given by :[tex]E=-L\dfrac{dI}{dt}[/tex][tex]E=-6.83\times 3[/tex]E = -20.49 voltHence, this is the required solution.

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