91影视

Faraday鈥檚 law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$蔚=-L\frac{di}{dt}$

Where L is the inductance of the conductor.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

Lenz further explained the direction of this induced current. According to lens, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

Number of turns, N = 500

Cross-sectional area, A = 6.25 cm

Mean Radius of the coil, r = 4 cm

Self-Inductance of the coil is given by

$L=\frac{{渭}_0{N}^{2}A}{2蟺r}$

$$\begin{gathered} =\frac{\left(4蟺*{10}^{-7}Tm/A\right)\left({500}^2\right)\left(6.25*{10}^{-4}{m}^2\right)}{2蟺\left(0.04m\right)} \\ =7.81x{10}^{-4} \\ =0.781mH \end{gathered}$$

Therefore, the self-inductance of the coil is 0.781 milli-henry.

Change in current,鈪�= 5.00 A 鈥� 2.00 A = 3.00 A

Change in time,螖t= 3.00 sec

Rate of change of current is given by

$$\begin{gathered} \frac{di}{dt}=\frac{鈭�i}{鈭�t} \\ =\frac{3\mathrm{A}}{3*{10}^{-3}\sec } \\ =1000\mathrm{A}/\sec \end{gathered}$$

Thus, induced emf is given by

$$\begin{gathered} 蔚=-L\frac{di}{dt} \\ =-\left(7.81x{10}^{-4H}\right)\left(1000\mathrm{A}/\sec \right) \\ =0.781V \end{gathered}$$

Therefore, the self-induced emf of the coil is 0.781 volts.

As the current is decreasing while moving from a to b, the direction of the induced will be acting in opposite direction of the current i.e., from b to a which means the terminal a is at higher potential than terminal b.