Question: Q. 2. An electron of mass $m_{e}$ revolves around a nucleus of charge $+Z_{e}$. Show that it behaves like a tiny magnetic dipole. Hence, prove that the magnetic moment associated with it is expressed as $\vec{\mu}=-\frac{e}{2 m_{e}} \vec{L}$, where $\vec{L}$ is the orbital angular momentum of the electron. Give the significance of negative sign.
U [Delhi II 2017]
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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,
$$ \begin{equation*} \therefore \quad \frac{R_{1}}{R_{2}}=\frac{\frac{8}{3}}{4}=\frac{2}{3} \tag{1} \end{equation*} $$