Physics 3 Marks Question

1. At a point just inside the tube the current enclosed in the closed surface = 0.

Thus $\text{B}=\frac{\mu_0\text{o}}{\text{A}}=0$

2. Taking a cylindrical surface just out side the tube, from ampere’s law

$\mu_0\text{i}=\text{B}\times2\pi\text{b}$

$\Rightarrow\text{B}=\frac{\mu_0\text{i}}{2\pi\text{b}}$

Force acting on 10cm of wire is 2 ×10^-5N

$\frac{\text{dF}}{\text{dl}}=\frac{\mu_0\text{i}_1\text{i}_2}{2\pi\text{d}}$

$\Rightarrow\frac{2\times10^{-5}}{10\times10^{-2}}=\frac{\mu_0\times20\times20}{2\pi\text{d}}$

$\Rightarrow\text{d}=\frac{4\pi\times10^{-7}\times20\times20\times10\times10^{-2}}{2\pi\times2\times10^{-5}}$

$=400\times10^{-3}=0.4\text{m}=40\text{cm}$

$\text{B}=\frac{\mu_0\text{i}}{2\text{R}}\frac{\theta}{2\pi}=\frac{2\pi}{3\times2\pi}\times\frac{\mu_0\text{i}}{2\text{R}}$

$=\frac{4\pi\times10^{-7}\times6}{6\times10^{-2}}=4\pi\times10^{-6}$

$=4\times3.14\times10^{-6}=12.56\times10^{-6}$

$=1.26\times10^{-5}\text{T}$

Given:

Magnitude of charges, q = 3.14 × 10⁻⁶C

Radius of the ring, r = 20cm = 20 × 10^-2mr = 20cm = 20 × 10^-2m

Angular velocity of the ring, $\omega=60\ \text{rad/s}$

Time for 1 revolution $=\frac{2\pi}{60}$

$\therefore\ \text{Current},\ \text{i}=\frac{\text{q}}{\text{t}}=\frac{3.14\times10^{-6}\times60}{2\pi}$

$=30\times10^{-6}\text{A}$

In the figure, E₁ and E₂ denotes the electric field at a point on the axis at a distance of 5.00 cm from the centre due to small element 1 and 2 of the ring respectively.

E is the resultant electric field due to the entire ring at a point on the axis at a distance of 5.00 cm from the centre.

The electric field at a point on the axis at a distance x from the centre is given by

$\text{E}=\frac{\text{xq}}{4\pi\in_0(\text{x}^2+\text{r}^2)^\frac{3}{2}}$

The magnetic field at a point on the axis at a distance x from the centre is given by

$\text{B}=\frac{\mu_0}{2}\frac{\text{ir}^2}{(\text{x}^2+\text{r}^2)^\frac{3}{2}}$

$\frac{\text{E}}{\text{B}}=\frac{\frac{\text{xq}}{4\pi\in_0(\text{x}^2+\text{r}^2)^\frac{3}{2}}}{\frac{\mu_0}{2}\frac{\text{ir}^2}{(\text{x}^2+\text{r}^2)^\frac{3}{2}}}$

$=\frac{9\times10^9\times3.14\times10^{-6}\times2\times(20.6)^3\times10^{-6}}{25\times10^{-4}\times4\pi\times12}$

$=\frac{9\times3.14\times2\times(20.6)^3}{25\times4\pi\times12}$

$=1.88\times10^{15}\text{m/s}$

The force acting on the smaller loop

$\text{F}=\text{ilB}\sin\theta$

$=\frac{\text{i}2\pi\text{r}\mu_0\text{I}1}{2\text{R}\times2}=\frac{\mu_0\text{iI}\pi\text{r}}{2\text{R}}$

$\text{F}_\text{AB} =\text{F}_\text{CD} +\text{F}_\text{EF}$

$=\frac{\mu_0\times10\times10}{2\pi\times10\times10}+\frac{\mu_0\times10\times10}{2\pi\times2\times10^{-2}}$

$=2\times10^{-3}+10^{-3}=3\times10^{-3}$ downward

$\text{F}_\text{CD} =\text{F}_\text{AB} +\text{F}_\text{EF}$

As F_AB & F_EF are equal and oppositely directed hence F = 0.

1. The maximum magnetic field is $\text{B}+\frac{\mu_0\text{I}}{2\pi\text{r}}$ which are along the left keeping the sense along the direction of traveling current.

2. The minimum $\text{B}-\frac{\mu_0\text{I}}{2\pi\text{r}}$

If $\text{r}=\frac{\mu_0\text{I}}{2\pi\text{B}}\ \text{B}\ \text{net}=0$

$\text{r}<\frac{\mu_0\text{I}}{2\pi\text{B}}\ \text{B}\ \text{net}=0$

$\text{r}>\frac{\mu_0\text{I}}{2\pi\text{B}}\ \text{B}\ \text{net}=\text{B}-\frac{\mu_0\text{I}}{2\pi\text{r}}$

For AB B is along $\odot\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ+\sin60^\circ)$

For AC B $\otimes\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ+\sin60^\circ)$

For BD B $\odot\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ)$

For DC $\otimes\ \text{B}=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin60^\circ)$

$\therefore\ \text{Net}\ \text{B}=0$

$\cos\theta=\frac{1}{2},$

$\theta=60^\circ\ \&\ \angle\text{AOB}=60^\circ$

$\text{B}=\frac{\mu_0\text{I}}{2\pi\text{r}}=\frac{10^{-7}\times2\times10}{2\times10^{-2}}=10^{-4}\text{T}$

So net is $[(10^{-4})^2]+(10^{-4})^2+2(10^{-8})\cos60^\circ]^\frac{1}{2}$

$=10^{-4}\Big[1+1+2\times\frac{1}{2}\Big]^\frac{1}{2}$

$=10^{-4}\times\sqrt3\text{T}$

$=1.732\times10^{-4}\text{T}$

1. For inside the tube $\overrightarrow{\text{B}}=0$

As, $\overrightarrow{\text{B}}$ inside the conducting tube = o

2. For $\overrightarrow{\text{B}}$ outside the tube

$\text{d}=\frac{3\text{r}}{2}$

$\overrightarrow{\text{B}}=\frac{\mu_0\text{i}}{2\pi\text{d}}=\frac{\mu_0\text{i}\times2}{2\pi3\text{r}}=\frac{\mu_0\text{i}}{2\pi\text{r}}$

$\overrightarrow{\text{B}}$ due to loop $\frac{\mu_0\text{i}}{2\text{r}}$

Let the straight current carrying wire be kept at a distance R from centre. Given I = 4i

$\overrightarrow{\text{B}}$ due to wire $\frac{\mu_0\text{i}}{2\pi\text{R}}=\frac{\mu_0\times4\text{i}}{2\pi\text{R}}$

Now, the $\overrightarrow{\text{B}}$ due to both will balance each other

Hence $\frac{\mu_0\text{i}}{2\text{r}}=\frac{\mu_04\text{i}}{2\pi\text{R}}\Rightarrow\text{R}=\frac{4\text{r}}{\pi}$

Hence the straight wire should be kept at a distance $\frac{4\pi}{\text{r}}$ from centre in such a way that the direction of current in it is opposite to that in the nearest part of circular wire. As a result the direction will $\overrightarrow{\text{B}}$ will be oppose.