Answer to Question #212009 in Electricity and Magnetism for Hetisani Sewela

Question

3. (a) Calculate the energy stored in a solenoid of self-inductance 0.1 H when a steady current of 1A is flowing through it? (5)

(b) What is the reason that a self-inductance cannot store energy for long periods of time. (5)

4. Two long parallel wires are separated by 100 cm, and each carry 1 A current in the same direction. Calculate the force between the wires per unit length. (µ_o = 1.26x10^-6Wb A^-1 m^-2) [10]

Answer 1

3.

(a)

Inductance

L = 0.1 H

Current

I = 1 A

Stored Energy $= 0.5 \times L \times I^2$

$= 0.5 \times 0.1 \times 1^2 \\ = 0.05 \;J \\ = 50\; mJ$

(b)

Inductor stores energy in form of magnetic field which is caused due to the steady current flow through the inductor. In case of a induction with zero resistance the energy will be stored for infinite amount of time. But as every wire has a resistor value thus due to the current flow through the inductor there will be a IR drop (heat generation). This is the leakage energy. Due to this leakage energy self-inductance inductors cannot store energy for long periods of time.

4. Force between the wires per unit length

$\frac{ΔF}{ΔL}= \frac{\mu_0I_1I_2}{2\pi d} \\ d=100 \;cm = 100 \times 10^{-2} \;m \\ \frac{ΔF}{ΔL}= \frac{1.26 \times 10^{-6} \times 1 \times 1}{2 \times 3.14 \times 100 \times 10^{-2}}= 2.006 \times 10^{-7} \; N/m$