Question: Q. 1. A rod of length $L$, along East West direction is dropped from a height $H$. If $B$ be the magnetic field due to earth at that place and angle of dip is $\theta$, then what is the magnitude of induced emf across two ends of the rod when the rod reaches the earth.
U] [CBSE SQP, 2016]
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Solution:
Ans.
$$ \begin{align*} \varepsilon & =\mathrm{B} l v \ & =B \cos \theta \times \mathrm{L} \times(2 \mathrm{gH})^{1 / 2} \tag{1} \end{align*} $$
[CBSE Marking Scheme, 2016]
Detailed Answer :
If a rod of length $\mathrm{L}$ is dropped from the height $\mathrm{H}$, it will cut the horizontal component of earth’s magnetic field thereby changing the magnetic flux by inducing an emf. When $B$ is the magnetic field and $\theta$ is the angle of dip, then horizontal magnetic field will be $B \cos \theta$ and velocity of rod dropped will be $\sqrt{2 g H}$ Hence, induced emf will be :
$$ \begin{align*} & |e|=B \cos \theta \times L \times v \ & |e|=B \cos \theta \times L \times \sqrt{2 g H} \tag{1} \end{align*} $$
AI Q. 2. An electron in an atom revolves around the nucleus in an orbit of radius $r$ with frequency $v$. Write the expression for the magnetic moment of the electron.
A [Foreign III 2014]
Ans. An electron revolving around the nucleus in an orbit will behave as current loop, so it will possess magnetic momentum,
$$ M=I A $$
where,
$I=$ current flowing through circular loop
Now, $\quad I=\frac{e}{T}$ where, $T=$ time period $=\frac{1}{v}$
$$ \text { So, } \quad \frac{1}{T}=v $$
The area of circular loop $A=\pi \mathrm{r}^{2}$
Hence from above, $M=e v\left(\pi r^{2}\right)$
[CBSE Marking Scheme 2014]
