Solution:

Ans. (i) Behaviour of revolving electron as a tiny magnetic dipole.

(ii) Proof of the relation, $\vec{\mu}=-\frac{e}{2 m_{e}} \vec{L}$

(iii) Significance of negative sign

Electron, in circular motion around the nucleus constitutes a current loop which behaves like a magnetic dipole.

Current associated with the revolving electron

$$ \begin{aligned} I & =\frac{e}{T} \ \text { and } \quad T & =\frac{2 \pi r}{v} \ \therefore \quad I & =\frac{-e}{2 \pi r} v \end{aligned} $$

Magnetic moment of the loop,

$$ \begin{aligned} \mu & =I A \ \mu & =I A=\frac{-e v}{2 \pi r} \pi r^{2} \ & =\frac{-e v r}{2}=\frac{-e \cdot m_{e} v r}{2 m_{e}} \end{aligned} $$

Orbital angular momentum of the electron,

-ve sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.

[CBSE Marking Scheme 2017]

[AI Q. 3. Define the term current sensitivity of a galvanometer. In the circuits shown in the figures, the galvanometer shows no deflection in each case. Find the ratio of $R_{1}$ and $R_{2}$.

U] [O.D. Comptt I, III 2017]

Ans. Definition of current sensitivity

Ratio $\frac{R_{1}}{R_{2}}$

Current sensitivity of a galvanometer is deflection per unit current

$$ \begin{array}{ll} \text { [Alternatively, } I_{s}=\frac{\theta}{I}=\frac{N A B}{K} \text { ] } \ \text { In circuit (i), } \frac{4}{6}=\frac{R_{1}}{4} \Rightarrow R_{1}=\frac{8}{3} \Omega \ \text { In circuit (ii), } \frac{6}{R_{2}}=\frac{12}{8} \Rightarrow R_{2}=4 \Omega \ \therefore \quad \frac{R_{1}}{R_{2}}=\frac{2}{3} \end{array} $$

[CBSE Marking Scheme 2017]

Detailed Answer :

(i) Current sensitivity of galvanometer is the deflection produced when unit current passes through the galvanometer. A galvanometer is said to be sensitive if it produces large deflection for a small current.

$$ I=\frac{C}{n B A} \theta $$

where, $C$ is torsional constant.

$$ \text { Currentsensitivity }=\frac{\theta}{I}=\frac{n B A}{C} $$

(ii) Consider the circuit I :

For balanced Wheatstone bridge, there will be no deflection in the galvanometer, so

Hence,

$$ \begin{aligned} \frac{4}{R_{1}} & =\frac{6}{4} \ \mathrm{R}_{1} & =\frac{4 \times 4}{6}=\frac{8}{3} \Omega \end{aligned} $$

$1 / 2$

Consider the circuit II :

For equivalent circuit, when the Wheatstone bridge is balanced, there will be no deflection in the galvanometer, so

Hence,

$$ \frac{6}{12}=\frac{R_{2}}{8} $$

$$ \mathrm{R}_{2}=\frac{6 \times 8}{12}=4 \Omega $$

Now ratio,

$$ \begin{equation*} \therefore \quad \frac{R_{1}}{R_{2}}=\frac{\frac{8}{3}}{4}=\frac{2}{3} \tag{1} \end{equation*} $$