Question

A long, straight solenoid with a cross-sectional area of 8.00cm2 is wound with 90turns of wire per centimeter, and the windings carry a current of 0.350A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400s.
What is the average induced emf in the second winding? Give your answer in Volts in scientific notation to three significant digits.

Answer 1

Answer:

the average induced emf in the second winding is $7.9168*10^{-5}V$

Explanation:

The magnetic field inside the first solenoid is given by,

$B= \mu_0NI$

Where $\mu$  is the permeability of the free space

N is the number of turns of solenoid per unit length

I is the current in the solenoid

A is the cross-sectional area of the wire

Replacing we have,

$B= (4*\pi*10^{-7})(90/cm (\frac{100cm}{1m}))(0.350A)$

$B = 3.9584*10^{-3}T$

Thus average emf induced in the second windigs is,

$\epsilon{avg}=\N\frac{dB}{dt}$

$\epsilon_{avg}=\frac{d}{dt}(AB)$

$\epsilon_{avg}=A\frac{dB}{dt}$

$\epsilon_{avg}= A\frac{dB}{dt}$

$\epsilon_{avg}=(8*10^{-4})\frac{3.9584*10^{-3}}{0.04}$

$\epsilon_{avg} = 7.9168*10^{-5}V$

Therefore the average induced emf in the second winding is $7.9168*10^{-5}V$