Magnetic Effect of Current MCQ Questions & Answers in Electrostatics and Magnetism | Physics

The force acting on electron will be perpendicular to the direction of velocity till the electron remains in the magnetic field. So the electron will follow the path as given.

A galvanometer can be converted into an ammeter by using a low resistance wire in parallel with the galvanometer. The resistance of this wire depends upon the range of the ammeter and it is given by

Shunt resistance $$S = \left( {\frac{{{i_g}G}}{{i - {i_g}}}} \right)$$

Apply Ampere's circular law to the coaxial circular loops $${L_1}$$ and $${L_2}.$$ The magnetic field is $${B_1}$$ at all points on $${L_1}$$ and $${B_2}$$ at all points on $${L_2}.$$ $$\sum I \ne 0$$   for $${L_1}$$ and 0 for $${L_2}.$$
Hence, $${B_1} \ne 0$$  but $${B_2} = 0$$
$$\left[ {{\text{As}}\,\oint {\vec B.d\vec i = {\mu _0}\sum I} } \right]$$

The angular momentum $$L$$ of the particle is given by $$L = m{r^2}\omega \,{\text{where}}\,\,\omega = 2\pi n.$$
$$\eqalign{ & \therefore {\text{Frequency}}\,n = \frac{\omega }{{2\pi }};{\text{Further}}\,i = q \times n = \frac{{\omega q}}{{2\pi }} \cr & {\text{Magnetic}}\,{\text{moment,}}\,M = iA = \frac{{\omega q}}{{2\pi }} \times \pi {r^2}; \cr & \therefore M = \frac{{\omega q{r^2}}}{2}\,{\text{so,}}\,\frac{M}{L} = \frac{{\omega q{r^2}}}{{2m{r^2}\omega }} = \frac{q}{{2m}} \cr} $$

Magnetic field due to a long current carrying wire at distance $$r$$ at point $$P$$ is given by
$$B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{r}$$
So, $$B \propto \frac{1}{r}$$
when $$r$$ is doubled, the magnetic field becomes halved.
i.e., $$B' = \frac{B}{2} \Rightarrow B' = \frac{{0.4}}{2} = 0.2\,T$$

Magnetic force = centripetal force
i.e. $$qvB = \frac{{m{v^2}}}{r}$$
or $$qvB = mr{\omega ^2}\,\,\left( {v = r\omega } \right)$$
or $${\omega ^2} = \frac{{qvB}}{{mr}} = \frac{{q\left( {r\omega } \right)B}}{{mr}}$$
Angular frequency, $$\omega = \frac{{qB}}{m}$$
If $$\nu $$ is the frequency of rotation, then
$$\eqalign{ & \omega = 2\pi \nu \Rightarrow \nu = \frac{\omega }{{2\pi }} \cr & \therefore \nu = \frac{{qB}}{{2\pi m}} \cr} $$
NOTE
In the resultant expression $$\frac{q}{m}$$ is known as specific charge. It is sometimes denoted by $$\alpha .$$ So, in terms of $$\alpha ,$$ the above formula can be written as
$$\omega = B\alpha \,\,{\text{and}}\,\,\nu = \frac{{B\alpha }}{{2\pi }}$$

As, $$I = \frac{q}{t}.$$  So, for an electron revolving in a circular orbit of radius $$r$$

$$q = e\,\,{\text{and}}\,\,t = T$$
$$ \Rightarrow I = \frac{e}{T} = \frac{e}{{\frac{{2\pi }}{\omega }}} = \frac{{\omega e}}{{2\pi }} = \frac{{2\pi ne}}{{2\pi }} = ne$$
The magnetic field produced at the centre is
$$B = \frac{{{\mu _0}I}}{{2R}} = \frac{{{\mu _0}ne}}{{2r}}$$

$${B_L} = \frac{{{\mu _0}I}}{{2\pi R}}$$
If the radius is $$\frac{R}{n};$$  the number of turns will be $$n.$$
$$\eqalign{ & {B_C} = \frac{{n{\mu _0}I}}{{2\pi \left( {\frac{R}{n}} \right)}} = {n^2}\frac{{{\mu _0}I}}{{2\pi R}} \cr & {\text{Hence,}}\,\,\frac{{{B_L}}}{{{B_C}}} = \frac{1}{{{n^2}}} \cr} $$

The situation is shown in the figure.
$${F_E}$$ = Force due to electric field
$${F_B}$$ = Force due to magnetic field
It is given that the charged particle remains moving along $$X$$-axis (i.e. undeviated). Therefore $${F_B} = {F_E}$$
$$ \Rightarrow qvB = qE \Rightarrow B = \frac{E}{v} = \frac{{{{10}^4}}}{{10}} = {10^3}\,weber/{m^2}$$

Magnetic force between parallel wires per unit length is given by $$\frac{F}{l} = \frac{{{\mu _0}}}{{2\pi }} \times \frac{{{i_1}{i_2}}}{r}$$
where, $${{i_1}}$$ and $${{i_2}}$$ are the currents in wires 1 and 2 respectively and $$r$$ is the distance between them.
Since, it is given that between two wires there is a force of attraction, so, the direction of currents in both will be the same.
$$\eqalign{ & {\text{Given,}}\,{i_1} = {i_2} = 1A,r = 1\,m, \cr & {\mu _0} = 4\pi \times {10^{ - 7}}T{\text{ - }}m/A \cr & \therefore \frac{F}{l} = \frac{{4\pi \times {{10}^{ - 7}}}}{{2\pi }} \times \frac{{1 \times 1}}{1} \cr & = 2 \times {10^{ - 7}}N/m \cr} $$
NOTE
When current is in same direction in both the wires there will be attraction and if current in opposite direction there is repulsion.