Physics 5 Marks Questions

1. The particulars will not collide if,

$\text{d}=\text{r}_1+\text{r}_2$

$\Rightarrow\text{d}=\frac{\text{mV}_\text{m}}{\text{qB}}+\frac{\text{mV}_\text{m}}{\text{qB}}$

$\Rightarrow\text{d}=\frac{2\text{mV}_\text{m}}{\text{qB}}$

$\text{V}_\text{m}=\frac{\text{qBd}}{2\text{m}}$

2. $\text{V}=\frac{\text{V}_\text{m}}{2}$

$\text{d}_1'=\text{r}_1+\text{r}_2=\Big(\frac{\text{m}\times\text{qBd}}{2\times2\text{m}\times\text{qB}}\Big)=\frac{\text{d}}{2}$ (min. dist.)

Max. distance $\text{d}_2'=\text{d}+2\text{r}=\text{d}+\frac{\text{d}}{2}=\frac{3\text{d}}{2}$

3. $\text{V}=2\text{V}_\text{m}$

$\text{r}_1'=\frac{\text{m}_2\text{V} _\text{m}}{\text{qB}}=\frac{\text{m}\times2\times\text{qBd}}{2\text{n}\times\text{qB}}$

$\text{r}_2=\text{d}$

$\therefore$ The arc is $\frac{1}{6}$

4. $\text{V}_\text{m}=\frac{\text{qBd}}{2\text{m}}$

The particles will collide at point P. At point p, both the particles will have motion m in upward direction. Since the particles collide inelastically the stick together.

Distance l between centres $=\text{d},\sin\theta=\frac{1}{2\text{r}}$

Velocity upward $=\text{v}\cos90-\theta=\text{V}\sin\theta=\frac{\text{Vl}}{2\text{r}}$

$\frac{\text{mv}^2}{\text{r}}=\text{qvB}$

$\Rightarrow\text{r}=\frac{\text{mv}}{\text{qB}}$

$\text{V}\sin\theta=\frac{\text{vl}}{2\text{r}}=\frac{\text{vl}}{2\frac{\text{mv}}{\text{qb}}}=\frac{\text{qBd}}{2\text{m}}=\text{V}_\text{m}$

Hence the combined mass will move with velocity V_m.

$\text{Y}=\frac{\text{Stress}}{\text{Strain}}=\frac{\Big(\frac{\text{F}}{\pi\text{r}^2}\Big)}{\Big(\frac{\text{dl}}{\text{L}}\Big)}$

$\Rightarrow\frac{\text{dl}}{\text{L}}\text{Y}=\frac{\text{F}}{\pi\text{r}^2}$

$\Rightarrow\text{dl}=\frac{\text{F}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{Y}}$

$=\frac{\text{iaB}}{\pi\text{r}^2}\times\frac{2\pi\text{a}}{\text{Y}}=\frac{2\pi\text{a}^2\text{iB}}{\pi\text{r}^2\text{Y}}$

So, $\text{dp}=\frac{2\pi\text{a}^2\text{iB}}{\pi\text{r}^2\text{Y}}$ (for small cross sectional circle)

$\text{dr}=\frac{2\pi\text{a}^2\text{iB}}{\pi\text{r}^2\text{Y}}\times\frac{1}{2\pi}=\frac{\text{a}^2\text{iB}}{\pi\text{r}^2\text{Y}}$

Considering a strip of width dx at a distance x from centre,

$\text{di}\frac{\text{dp}}{\text{dt}}=\frac{\text{q}4\pi\text{x}^2\text{dx}}{\Big(\frac{4}{3}\Big)\pi\text{R}^3\text{}}4\pi\text{x}^2\text{dx}$

$\text{di}=\frac{\text{dq}}{\text{dt}}=\frac{\text{q}4\pi\text{x}^2\text{dx}}{\Big(\frac{4}{3}\Big)\pi\text{R}^3\text{t}}=\frac{3\text{q}\text{x}^2\text{dx}\omega}{\text{R}^32\pi}$

$\text{d}\mu=\text{di}\times\text{A}=\frac{3\text{q}\text{x}^2\text{d}\text{x}\omega}{\text{R}^32\pi}\times4\pi\text{x}^{2\ =\ \frac{6\text{q}\omega}{\text{R}^3}\text{x}^4\text{dx}}$

$\mu=\int\limits_0^{\mu}\text{d}\mu\int\limits_0^{\text{R}}\frac{6\text{q}\omega}{\text{R}^3}\text{x}^4\text{dx}=\frac{6\text{q}\omega}{\text{R}^3}\Big[\frac{\text{x}^5}{5}\Big]_0$

$\text{R}=\frac{\text{6}\text{q}\omega\text{R}^5}{\text{R}^3}\frac{\text{R}^5}{5}=\frac{6}{5}\text{q}\omega\text{R}^2$

$\overrightarrow{\text{B}}=\text{B}_0\Big(1+\frac{\text{x}}{\text{l}}\Big)\hat{\text{k}}$

f₁ = force on AB $=\text{iB}_0[1+0]\text{l}=\text{iB}_0\text{l}$

f₂ = force on CD$=\text{iB}_0[1+0]\text{l}=\text{iB}_0\text{l}$

f₃ = force on AD$=\text{iB}_0\Big[1+\frac{0}{1}\Big]\text{l}=\text{iB}_0\text{l}$

f₄ = force on AB $=\text{iB}_0\Big[1+\frac{1}{1}\Big]2=\text{iB}_0\text{l}$

Net horizontal force $=​​​​​​\text{F}_1-\text{F}_2=0$

Net vertical force$=\text{F}_4-\text{F}_3=\text{iB}_0\text{l}$

1. Mass of the particle = m, Charge = q, Width = d

if $\text{d}=\frac{\text{mV}}{\text{qB}}$

The d is equal to radius. $\theta$ is the angle between the radius and tangent which is equal to $\frac{\pi}{2}$ (As shown in the figure)

2. if $\approx\frac{\text{mV}}{2\text{qB}}$ distance travelled $=\Big(\frac{1}{2}\Big)$ of radius

Along x-directions d = V_Xt [Since acceleration in this direction is 0. Force acts along $\overrightarrow{\text{j}}$ directions]

$\text{t}=\frac{\text{d}}{\text{V}_\text{X}}\ ...(1)$

$\text{V}_\text{Y}=\text{u}_\text{Y}+\text{a}_\text{Y}\text{t}=\frac{0+\text{qu}_\text{X}\text{Bt}}{\text{m}}=\frac{\text{qu}_\text{Y}\text{Bt}}{\text{m}}$

From (1) putting the value of t, $\text{V}_\text{Y}=\frac{\text{qu}_\text{X}\text{Bd}}{\text{m}\text{V}_\text{X}}$

$\text{Tan}\theta\frac{\text{V}_\text{Y}}{\text{V}_\text{X}}=\frac{\text{qBd}}{\text{mV}_\text{X}}=\frac{\text{qBmV}_\text{X}}{2\text{qBmV}_\text{X}}=\frac{1}{2}$

$\Rightarrow\theta=\tan^{-1}\Big(\frac{1}{2}\Big)=26.4\approx30^\circ=\frac{\pi}{6}$

3. $\text{d}\approx\frac{2\text{mu}}{\text{qB}}$

Looking into the figure, the angle between the initial direction and final direction of velocity is $\pi. $

l = 20cm = 20 × 10^-2m

B = 10cm = 10 × 10^-2m

i = 5A,

B = 0.2T

There is no force on the sides AB and CD. But the force on the sides AD and BC are opposite. So they cancel each other.

Torque on the loop

$\tau=\text{ni}\overrightarrow{\text{A}}\times\overrightarrow{\text{B}}=\text{ni}\text{ AB}\sin90^\circ$

$\Rightarrow1\times5\times20\times10^{-2}\times10\times10^{-2}0.2$

$=2\times10^{-2}=0.02\text{N-M}$

Parallel to the shorter side.

$\text{u}\times10^4\text{m/s}$

$\text{B}=0.5\text{T},$

$\text{r}=\frac{3}{2}=1.5\text{cm},$

$\text{r}_1=\frac{\text{mv}}{\text{qB}}=\frac{\text{A}\times(1.6\times10^{27})\times6\times10^4}{1.6\times10^{-19}\times0.5}$

$\Rightarrow1.5=\text{A}\times12\times10^{-4}$

$\Rightarrow\text{A}=\frac{1.5}{12\times106{-4}}=\frac{15000}{12}$

$\text{r}_2=\frac{\text{mu}}{\text{qB}}$

$\Rightarrow\frac{3.5}{2}=\frac{\text{A}'\times(1.6\times106{-27})\times6\times10^4}{1.6\times10^{-19}\times0.5}$

$\Rightarrow\text{A}'=\frac{3.5\times0.5\times10^{-19}}{2\times6\times10^4\times10^{-27}}$

$=\frac{3.5\times0.5\times10^4}{12}$

$\frac{\text{A}}{\text{A}'}=\frac{1.5}{12\times10^{-4}}\times\frac{12\times10^{-4}}{3.5\times0.5}=\frac{6}{7}$

Taking common ration = 2 (For Carbon). The isotopes used are C¹² and C¹⁴

The force experienced first is due to the electric field due to the capacitor

$\text{E}=\frac{\text{V}}{\text{d}}$

$\text{F}=\text{eE}$

$\text{a}=\frac{\text{eE}}{\text{m}_\text{e}}$ [Where e → charge of electron m_e → mass of electron]

$\text{v}^2=\text{u}^2+2\text{ as}$

$\Rightarrow\text{v}^2=2\times\frac{\text{eE}}{\text{m}_\text{e}}\times\frac{2\times\text{e}\times\text{V}\times\text{d}}{\text{dm}_\text{e}}$

$\text{v}=\sqrt{\frac{2\text{eV}}{\text{m}_\text{e}}}$

Now, The electron will fail to strike the upper plate only when d is greater than radius of the are thus formed.

$\text{d}>\frac{\text{m}_\text{e}\times\sqrt{\frac{2\text{eV}}{\text{m}_\text{e}}}}{\text{eB}}$

$\Rightarrow\text{d}>\frac{\sqrt{2\text{m}_\text{e}\text{V}}}{\text{eB}^2}$

dp on the small length dx is $\frac{\text{q}}{\pi\text{r}^2}2\pi\text{x}\text{ dx}.$

$\text{di}=\frac{\text{q}2\pi\times\text{dx}}{\pi\text{r}^2\text{t}}=\frac{\text{q}2\pi\text{dx}\omega}{\pi\text{r}^2\text{q}2\pi}=\frac{\text{q}\omega}{\pi\text{r}^2}\text{xdx}$

$\text{d}\mu=\text{n}\text{ di}\text{A}=\frac{\text{q}\omega\text{xdx}}{\pi\text{r}^2}\pi\text{x}^2$

$\mu=\int\limits_0^\mu\text{d}\mu=\int\limits_0^{\text{r}}\frac{\text{q}\omega}{\text{r}^2}\text{x}^3\text{dx}=\frac{\text{q}\omega}{\text{r}^2}\Big[\frac{\text{x}^4}{4}\Big]^\text{r}=\frac{\text{q}\omega\text{r}^4}{\text{r}^2\times4}=\frac{\text{q}\omega\text{r}^2}{4}$

$\text{l}=\text{I}\omega=\Big(\frac{1}{2}\Big)\text{mr}^2\omega$

$\Big[\therefore$ M.I. for disc is $\Big(\frac{1}{2}\Big)\text{mr}^2\Big]$

$\frac{\mu}{\text{l}}=\frac{\text{q}\omega\text{r}^2}{4\times\Big(\frac{1}{2}\Big)\text{mr}^2\omega}$

$\Rightarrow\frac{\text{q}}{\text{l}}=\frac{\text{q}}{2\text{m}}$

$\Rightarrow\mu=\frac{\text{q}}{2\text{m}}\text{l}$

Velocity will be along x - z plane

$\overrightarrow{\text{B}}-\text{B}_0\overrightarrow{\text{j}}$

$\overrightarrow{\text{E}}=\text{E}_0\hat{\text{k}}$

$\text{F}=\text{q}\big(\overrightarrow{\text{E}}+\overrightarrow{\text{V}}\times\overrightarrow{\text{B}}\big)$

$=\text{q}\Big[\text{E}_0\hat{\text{k}}+(\text{u}_\text{x}\hat{\text{i}}+\text{u}_\text{x}\hat{\text{k}})(-\text{B}_0\hat{\text{j}})\Big]$

$=(\text{qE}_0)\hat{\text{k}}-(\text{u}_\text{x}\text{B}_0)\hat{\text{k}}+(\text{u}_\text{z}\text{B}_0)\hat{\text{i}}$

$\text{F}_\text{z}=(\text{qE}_0-\text{u}_\text{x}\text{B}_0)$

since $\text{u}_\text{x}=0,\ \text{F}_\text{z}=\text{qE}_0$

$\Rightarrow\text{a}+\text{z}=\frac{\text{qE}_0}{\text{m}}$

So, $\text{v}^2=\text{u}^2+2$

as, $\Rightarrow\text{v}^2=2\frac{\text{qE}}{\text{m}}\text{Z}$ [distance along Z direction be z]

$\Rightarrow\text{V}\sqrt{\frac{2\text{qE}_0\text{Z}}{\text{m}}}$

$\text{KE}=10\text{Kev}=1.6\times10^{-15}\text{J}$

$\overrightarrow{\text{B}}=1\times10^{-7}\text{T}$

1. The electron will be deflected towards left
2. $\Big(\frac{1}{2}\Big)\text{mv}^2=\text{KE}$

$\Rightarrow\text{V}=\sqrt{\frac{\text{KE}\times2}{\text{m}}}$

$\text{F}=\text{qVB}\ \&\text{ accin}=\frac{\text{qVB}}{\text{m}_\text{e}}$

Applying $\text{s}=\text{ut}+\Big(\frac{1}{2}\Big)\text{at}^2=\frac{1}{2}\times\frac{\text{qVB}}{\text{m}_\text{e}}\times\frac{\text{x}^2}{\text{v}^2}$

$=\frac{\text{qBx}^2}{2\text{m}_\text{e}\text{V}}$

$=\frac{\text{qBx}^2}{2\text{m}_\text{e}\sqrt{\frac{\text{KE}\times2}{\text{m}}}}$

$=\frac{1}{2}\times\frac{1.6\times10^{-19}\times1\times10^{-7}\times1^2}{9.1\times10^{-31}\times\sqrt{\frac{1.6\times10^{-15}\times2}{9.1\times10^{-31}}}}$

By solving we get, $\text{s}=0.0148\approx1.5\times10^{-2}\text{cm}$

The object will make a circular path, perpendicular to the plance of paper Let the radius of the object be r.

$\frac{\text{mv}^2}{\text{r}}=\text{qvB}\Rightarrow\text{r}=\frac{\text{mV}}{\text{qB}}$

Here object distance K = 18cm.

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ (lens eqn.) $\Rightarrow\frac{1}{\text{v}}-\Big(\frac{1}{-18}\Big)=\frac{1}{12}\Rightarrow\text{v}=36\text{cm}.$

Let the radius of the circular path of image = r'

So magnification $=\frac{\text{v}}{\text{u}}=\frac{\text{r}'}{\text{r}}\Big(\text{magnetic path}=\frac{\text{image height}}{\text{object height}}\Big)$

$\Rightarrow\text{r}'=\frac{\text{v}}{\text{u}}\text{r}$

$\Rightarrow\text{r}'=\frac{36}{18}\times4=8\text{cm}$

Hence radius of the circular path in which the image moves is 8cm 

1. $\text{i}=\text{V}_0\text{n}\text{Ae}$

$\Rightarrow\text{V}_0=\frac{\text{i}}{\text{nae}}$

2. $\text{F}=\text{ilB}=\frac{\text{iB}}{\text{nA}}$ (upwards)
3. Let the electric field be E

$\text{Ee}=\frac{\text{iB}}{\text{An}}\Rightarrow\text{E}=\frac{\text{iB}}{\text{Aen}}$

4. $\frac{\text{dv}}{\text{dr}}=\text{E}\Rightarrow\text{dV}=\text{Edr}$

$=\text{E}\times\text{d}=\frac{\text{ib}}{\text{Aen}}\text{d}$

Velocity of electron = v

Magnetic force on electron

F = evB

2. F = qE ; F = evB

$\Rightarrow\text{qE}=\text{evB}$

$\Rightarrow\text{eE}=\text{evB}$

$\Rightarrow\overrightarrow{\text{E}}=\text{vB}$

3. $\text{E}=\frac{\text{dV}}{\text{dr}}=\frac{\text{V}}{\text{l}}$

$\Rightarrow\text{V}=\text{lE}=\text{lvB}$

1. Radius of circular arc $=\frac{\text{mv}}{\text{qB}}$
2. Since MA is tangent to are ABC, described by the particle.

Hence $\angle\text{MAO}=90^\circ$

Now, $\angle\text{NAC}=90^\circ[\because\text{NA}\text{ is}\perp\text{r}]$

$\therefore\angle\text{OAC}=\angle\text{OCA}=\theta$ [By geometry]

Then $\angle\text{AOC}=180-(\theta+\theta)=\pi-2\theta$

3. Dist. Covered $\text{l}=\text{r}\theta=\frac{\text{mv}}{\text{pB}}(\pi-2\theta)$

$\text{t}=\frac{\text{l}}{\text{v}}=\frac{\text{m}}{\text{qB}}(\pi-2\theta)$

4. If the charge ‘q’ on the particle is negative. Then

1. Radius of Circular arc $=\frac{\text{mv}}{\text{qB}}$
2. In such a case the centre of the arc will lie with in the magnetic field, as seen in the fig. Hence the angle subtended by the major arc $=\pi+2\theta$
3. Similarly the time taken by the particle to cover the same path $=\frac{\text{m}}{\text{qB}}(\pi+2\theta)$

$\overrightarrow{\text{F}}=\text{ilB}=1\times0.20\times0.1=0.02\text{N}$

For $\overrightarrow{\text{F}}=\text{il}\times\text{B}$

So, for

$\text{da}\ \&\ \text{cb}\rightarrow\text{l}\times\text{B}=\sin90^\circ$ towards left

Hence $\overrightarrow{\text{F}}0.02\text{N}$ towards left

For

$\text{dc}\ \&\ \text{ab}\rightarrow\overrightarrow{\text{F}}=0.02\text{N}$ downward

1. $\text{R}=1\text{n},$

$\text{B}=0.5\text{T},$

$\text{r}=\frac{\text{mv}}{\text{qB}}$

$\Rightarrow1=\frac{9.1\times10^{-31}\times\text{v}}{1.6\times10^{-19}\times0.5}$

$\Rightarrow\text{v}=\frac{1.6\times0.5\times106{-19}}{9.1\times10^{-31}}$

$=0.0879\times10^{10}\approx8.8\times10^{10}\text{m/s}$

No, it is not reasonable as it is more than the speed of light.

2. $\text{r}=\frac{\text{Mv}}{\text{qB}}$

$\Rightarrow1=\frac{1.6\times10^{27}\times\text{v}}{1.6\times10^{-19}\times0.5}$

$\Rightarrow\text{v}=\frac{1.6\times106{19}\times0.5}{1.6\times10^{27}}$

$=0.5\times10^8=5\times10^7\text{m/s}$

$\therefore\text{E}=\frac{\text{V}}{\text{R}'}$

$\theta$ = Sin^-1 (0.9),

$\text{r}=\frac{\text{mv}'}{\text{qB}}=\frac{\text{mv}\sin\theta}{\text{qB}}$

$=\frac{5\times10^{-12}\times10^3\times9}{5\times10^{-6}+5\times10^3+10}$

$=0.18\text{ metre}$

$\frac{\text{1}}{2}\text{mv}^2=32\times10^3\times1.6\times10^{-19}$

$\frac{1}{2}\times39\times(1.6\times10^{-27})\times\text{v}^2$

$=32\times10^3\times1.6\times10^{-19}$

$\text{t}=\frac{\text{d}}{\text{v}}=\frac{0.01}{4.05\times10^5}$

$\text{a}=\frac{\text{qvB}}{\text{m}}$

$=\frac{1.6\times10^{19}\times4.05\times10^5}{39\times1.6\times10^{-27}}$

$=\frac{\text{d}}{\text{v}}=\frac{0.955}{4.05\times10^5}$

$\frac{1}{2}\times41\times1.6\times10^{-27}$