Numerical Problems-2 - Chapter 3 Magnetism and Magnetic Effects of Electric Current 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
The radius of the first orbit of hydrogen atom is $0.5 \AA$. The electron moves in an orbit with a uniform speed of $2.2 \times 10^6 \mathrm{~ms}^{-1}$. What is the magnetic field produced at the centre of the nucleus due to the motion of this electron? Use $\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}$ and electric charge $=1.6 \times 10^{-19} \mathrm{C}$
Solution:
Here, $\mathrm{r}=0.5 \AA=0.5 \times 10^{-10} \mathrm{~m}$
$\mathrm{v}=2.2 \times 10^6 \mathrm{~ms}^{-1}$
Period of revolution of electron
$
\mathrm{T}=\frac{2 \pi r}{v}=\frac{2 \times 22 \times 0.5 \times 10^{-10}}{7 \times 2.2 \times 10^6}=\frac{1}{7} \times 10^{-15} \mathrm{~s}
$
Equivalent current, $\mathrm{I}=\frac{\text { charge }}{\text { Time }}=\frac{\mathrm{e}}{\mathrm{T}}=\frac{1.6 \times 10^{-19} \times 7}{10^{-15}}$ $=1.12 \times 10^{-3} \mathrm{~A}$
Magnetic field produced at the centre of the nucleus,
$
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 I}{2 r}=\frac{4 \pi \times 10^{-7} \times 1.12 \times 10^{-3}}{2 \times 0.5 \times 10^{-10}} \\
& \mathrm{~B}=14.07 \mathrm{~T}
\end{aligned}
$
Question 2.
A positive charge of $1.5 \mu \mathrm{C}$ is moving with a speed of $2 \times 10^{-6} \mathrm{~ms}^{-1}$ along the positive $\mathrm{X}$-axis. A magnetic field, $\vec{B}=(0.2 \hat{j}+0.4 \hat{k})$ tesla acts in space. Find the magnetic force acting on the charge. Solution:
Here $\mathrm{q}=1.5 \mu \mathrm{C}=1.5 \times 10^{-6} \mathrm{C}$
$
\vec{v}=2 \times 10^{-6} \hat{i} \mathrm{~ms}^{-1} ; \vec{B}=(0.2 \hat{j}+0.4 \hat{k}) \mathrm{T}
$
Magnetic force on the positive charge is
$
\begin{aligned}
& \vec{F}=(\vec{v} \times \vec{B}) \\
& =1.5 \times 10^{-6}\left[2 \times 10^6 \hat{i} \times(0.2 \hat{j}+0.4 \hat{k})\right] \\
& 3.0[0.2 \hat{j} \times \hat{j}+0.4 \hat{i} \times \hat{k}] \\
& =(0.6 \hat{k}-1.2 \hat{j}) \mathrm{N} \\
& {[\therefore \mathrm{i} \times \hat{j}=\hat{k}, \hat{i} \times \hat{k}=-\hat{j}]}
\end{aligned}
$
Question 3.
Copper has $8.0 \times 10^{28}$ electrons per cubic metre carrying a current and lying at right angle to a magnetic field of strength $5 \times 10^{-3} \mathrm{~T}$. experiences a force of $8.0 \times 10^{-2} \mathrm{~N}$. Calculate the drift velocity of free electrons in the wire.

Solution:
$
\begin{aligned}
& \mathrm{n}=8 \times 10^{28} \mathrm{~m}^{-3} ; 1=1 \mathrm{~m} \\
& \mathrm{~A}=8 \times 10^{28} \mathrm{~m}^{-2} ; \mathrm{e}=1.6 \times 10^{-9} \mathrm{C}
\end{aligned}
$
Total charge contained in the wire
$
\begin{aligned}
& \mathrm{q}=\text { volume of wire } \times \text { ne }=\text { Alne }=8 \times 10^{-6} \times 1 \times 8 \times 10^{28} \times 1.6 \times 10^{-19} \mathrm{C} \\
& =102.4 \times 10^3 \mathrm{C}
\end{aligned}
$
If $\mathrm{v}_{\mathrm{d}}$ is the drift speed of electrons, then
$
\begin{aligned}
& \mathrm{F}=\mathrm{qv}_{\mathrm{d}} \mathrm{B} \sin 90^{\circ}=\mathrm{qv}_{\mathrm{d}} \mathrm{B} \\
& \therefore \mathrm{v}_{\mathrm{d}}=\frac{F}{q B}=\frac{8.0 \times 10^{-2}}{102.4 \times 10^3 \times 5 \times 10^{-3}} \mathrm{~ms}^{-1} \\
& \mathrm{v}_{\mathrm{d}}=1.56 \times 10^{-4} \mathrm{~ms}^{-1}
\end{aligned}
$
Question 4.
An electron is moving at $10^6 \mathrm{~ms}^{-1}$ in a direction parallel to a current of $5 \mathrm{~A}$ flowing through an infinitely long straight wire, separated by a perpendicular distance of $10 \mathrm{~cm}$ in air. Calculate the magnitude of the force experienced by the electron.
Solution:
Magnetic field of the straight wire carrying a current of $2 \mathrm{~A}$, at a distance of $10 \mathrm{~cm}$ or $0.1 \mathrm{~m}$ from it is
$
\mathrm{B}=\frac{\mu_0 I}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.1}=10^{-5} \mathrm{~T}
$
This field acts perpendicular to the direction of the electron. So magnetic force on the electron is
$
\begin{aligned}
& \mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=1.6 \times 10^{-19} \times 10^6 10^{-5} \times 1 \\
& \mathrm{~F}=1.6 \times 10^{-18} \mathrm{~N}
\end{aligned}
$
Common Errors and Its Rectifications
Common Errors:
1. Students may not know about arrow mark in the figures, (diagram)
2. Students wrongly mention the unit of magnetic induction. Eg. $0.5 \times 10^{-5}$ Tesla. Most of the children makes the mistakes in this area.
3. Student may confuse the direction of current versus produced magnetic induction.
Rectifications:
1. Arrow mark is importance to show where the electric and magnetic fields are directed.
2. The correct unit of magnetic induction of $T$ (or) tesla.
3. If the current moves in upward the magnetic induction is anti-clock. If the current in downward, the magnetic induction is clockwise.