Question

A toroidal solenoid has 500 turns, cross-sectional area 6.25 cm2, and mean radius 4.00 cm. (a) Calculate the coil’s self-inductance. (b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil. (c) The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Short answer

a) The self-inductance of the coil is 0.781 milli-henry.

b) The self-induced emf of the coil is 0.781 volts.

c) Direction of emf is from b to a.

Step-by-step solution

Basic definition

Faraday’s 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

$\epsilon =-L\frac{di}{dt}$

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.

Self Inductance of the coil

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

$L=\frac{{\mu }_0{N}^{2}A}{2\pi r}$

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

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

Self Induced EMF

Change in current,ⅈ= 5.00 A – 2.00 A = 3.00 A

Change in time,Δt= 3.00 sec

Rate of change of current is given by

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

Thus, induced emf is given by

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

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

Direction of Induced EMF

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.