Solution:

Ans. Relative permeability

$$ \mu_{r}=\frac{L}{L_{0}} $$

$$ \begin{aligned} & \mu_{r}=\frac{2 \cdot 8}{2 \cdot 0 \times 10^{-3}} \quad 1 / 2 \ &=1400 \ & \text { BSE Marking Scheme 20171 } \end{aligned} $$

[CBSE Marking Scheme 2017]

AI Q. 2. A metallic piece gets hot when surrounded by a coil carrying high frequency alternating current. Why?

C] [Delhi Comptt. I, II, III 2014]

Ans. Due to the heating effect of eddy currents set up in the metallic piece. [CBSE Marking Scheme 2014] 1

Short Answer Type Questions-I

AI Q. 1. A light bulb and a solenoid are connected in series across an ac source of voltage. Explain, how the glow of the light bulb will be affected when an iron rod is inserted in the solenoid.

]] [Foreign I, II, III 2017]

Ans. Effect on brightness
Explanation

Brightness decreases

$1 / 2$ increse; this incr inductance of solenoid 侮 circuit and hence current decreases.

(Even student just writes self inductance increases, award this 1 mark.)

[CBSE Marking Scheme 2017]

Detailed Answer :

(i) Brightness will decrease when an iron rod is inserted in the solenoid.

(ii) When an iron rod is inserted in the solenoid with a velocity, the iron rod will cut the magnetic field lines of an inductor, so as per Kirchhoff’s and Faraday’s Laws, cutting of magnetic field will tend to induce a current inside the inductor which opposes the direction of it’s cause. So when the current is being induced by the moving rod, it opposes the flow of existing current in the circuit, causing the bulb’s brightness to go down as there is less current passing through it.

[AI Q. 2. Define self-inductance of a coil. Show that magnetic energy required to build up the current $I$ in a coil of self-inductance $L$ is given by $\frac{1}{2} L I^{2}$.

R[Delhi I, II, III 2014]

Ans. Self-Inductance is the property by which an opposing induced em is produced in a coil due to a change in current, or magnetic flux, linked with the coil.

OR

Self-inductance of a coil is numerically equal to the flux linked with the coil when the current through the coil is $1 \mathrm{~A}$.

$$ \text { OR } $$

Self-inductance of a coil is equal to the induced emf developed in the coil when the rate of change of current in the coil is one ampere per second. 1 Energy stored in an inductor:

Consider a source of emf connected to an inductor $L$. As the current starts growing, the opposing induced emf is given by

$$ \varepsilon=-\mathrm{L} \frac{d i}{d t} $$

If the source of emf flows a current $i$ through the inductor for a small time $d t$, then the amount of work done by the source, is given by $d W=|\varepsilon|$ idt

$$ \begin{aligned} & =L i \frac{d i}{d t} d t \ & =L i d i \end{aligned} $$

Hence the total amount of work done (by the source of emf) when the current increases from its initial value $(i=0)$ to its final value $(I)$ is given by

$W=\int_{0}^{\mathrm{I}} L i d i=L \int_{0}^{\mathrm{I}} i d i=L\left[\frac{i^{2}}{2}\right]_{0}^{\mathrm{I}}=\frac{1}{2} L I^{2}$

This work done gets stored in the inductor in the form of magnetic energy.

$\therefore \quad \quad U=\frac{1}{2} L I^{2}$

[CBSE Marking Scheme 2014]