153608
A proton and an $\alpha$-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. The ratio of the radii of their paths of proton to that of the $\alpha$-particle is
1 $2: 1$
2 $1: 2$
3 $1: 1$
4 $2: 3$
Explanation:
C From question, we know - Radius of circular path, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{Bq}}$ For same kinetic energy $\mathrm{K}$ and field $\mathrm{B}$, $r \propto \frac{\sqrt{m}}{q}$ $\text { So, } \quad \frac{r_{p}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{p}} \sqrt{\frac{m_{p}}{m_{\alpha}}}=\frac{2 e}{e} \sqrt{\frac{m_{p}}{4 m_{p}}}=\frac{1}{1}$ $r_{p}: r_{\alpha}=1: 1$
COMEDK 2014
Moving Charges & Magnetism
153609
What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, is required to make the electrons travel in a circular arc of radius $0.35 \mathrm{~m}$ ?
1 $2.1 \times 10^{-5} \mathrm{G}$
2 $6 \times 10^{-6} \mathrm{~T}$
3 $2.1 \times 10^{-5} \mathrm{~T}$
4 $6 \times 10^{-6} \mathrm{G}$
Explanation:
C Given that, $\mathrm{v}=1.3 \times 10^{6} \mathrm{~m} / \mathrm{sec}$ $\mathrm{r}=0.35 \mathrm{~m}$ Mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ Charge of electron $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ $\theta=90^{\circ}$ We know that, magnetic field in a circular arc, $\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qr}}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^{6}}{1.6 \times 10^{-19} \times 0.350}$ $\mathrm{~B}=\frac{11.83}{0.56} \times 10^{-6}$ $\mathrm{~B}=2.11 \times 10^{-5} \mathrm{~T}$
COMEDK 2016
Moving Charges & Magnetism
153610
A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be
1 $\mathrm{T}$
2 $4 \mathrm{~T}$
3 $3 \mathrm{~T}$
4 $2 \mathrm{~T}$
Explanation:
A When a charged particle enters in an uniform magnetic field, a magnetic force acts on the charged particle in the direction perpendicular to the direction of motion of the particle and this causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $90^{\circ}$. Then, $\quad \mathrm{W}=\mathrm{Fd} \cos 90^{\circ}$ $\mathrm{W}=0$ So, $\quad$ K.E. $=$ Work done $\text { K.E. }=0$ The change in kinetic energy of the charged particle after entering the magnetic field is zero. Hence, the kinetic energy of the charged particle after 3 second is $\mathrm{T}$.
COMEDK 2017
Moving Charges & Magnetism
153611
A magnetic field (B) perpendicular to plane of paper is present at a certain place. Calcium ion $\left(\mathrm{Ca}^{++}\right)$is moving in the plane with a velocity $v$. The magnetic force acting on it is
1 evB
2 $2 \mathrm{evB}$
3 $\frac{\text { evB }}{2}$
4 zero
Explanation:
B Magnetic force acting on a charge q moving with velocity $\overrightarrow{\mathrm{V}}$ in a magnetic field $\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \quad(\because \mathrm{a} \times \mathrm{b}=\mathrm{ab} \sin \theta)$ $\mathrm{F}=\mathrm{qvB} \sin \theta$ Where, $\mathrm{B}=$ flux density of magnetic field $\mathrm{q}=$ charge of the particle $\mathrm{v}=$ velocity of particle $\theta=$ angle between vector $\vec{v}$ and $\vec{B}$ For, charge on Calcium ion $\left(\mathrm{Ca}^{++}\right), \mathrm{q}=2 \mathrm{e}$ $\mathrm{F} =2 \mathrm{evB} \sin 90^{\circ}$ $\mathrm{F} =2 \mathrm{evB}$
COMEDK 2017
Moving Charges & Magnetism
153612
A charged particle moves through a magnetic field in a direction perpendicular to it. Then it
1 velocity remains unchanged
2 speed of the particle remains unchanged
3 direction of the particle remains unchanged
4 acceleration remains unchanged
Explanation:
B Given that, $\theta=90^{\circ}$ Force on a charged particle $\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{qvB} \sin \theta$ $\mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB}$ As force is acting perpendicular to velocity, this implies speed will not be changed only direction will be changed. Thus, speed of the particle remains unchanged.
153608
A proton and an $\alpha$-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. The ratio of the radii of their paths of proton to that of the $\alpha$-particle is
1 $2: 1$
2 $1: 2$
3 $1: 1$
4 $2: 3$
Explanation:
C From question, we know - Radius of circular path, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{Bq}}$ For same kinetic energy $\mathrm{K}$ and field $\mathrm{B}$, $r \propto \frac{\sqrt{m}}{q}$ $\text { So, } \quad \frac{r_{p}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{p}} \sqrt{\frac{m_{p}}{m_{\alpha}}}=\frac{2 e}{e} \sqrt{\frac{m_{p}}{4 m_{p}}}=\frac{1}{1}$ $r_{p}: r_{\alpha}=1: 1$
COMEDK 2014
Moving Charges & Magnetism
153609
What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, is required to make the electrons travel in a circular arc of radius $0.35 \mathrm{~m}$ ?
1 $2.1 \times 10^{-5} \mathrm{G}$
2 $6 \times 10^{-6} \mathrm{~T}$
3 $2.1 \times 10^{-5} \mathrm{~T}$
4 $6 \times 10^{-6} \mathrm{G}$
Explanation:
C Given that, $\mathrm{v}=1.3 \times 10^{6} \mathrm{~m} / \mathrm{sec}$ $\mathrm{r}=0.35 \mathrm{~m}$ Mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ Charge of electron $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ $\theta=90^{\circ}$ We know that, magnetic field in a circular arc, $\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qr}}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^{6}}{1.6 \times 10^{-19} \times 0.350}$ $\mathrm{~B}=\frac{11.83}{0.56} \times 10^{-6}$ $\mathrm{~B}=2.11 \times 10^{-5} \mathrm{~T}$
COMEDK 2016
Moving Charges & Magnetism
153610
A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be
1 $\mathrm{T}$
2 $4 \mathrm{~T}$
3 $3 \mathrm{~T}$
4 $2 \mathrm{~T}$
Explanation:
A When a charged particle enters in an uniform magnetic field, a magnetic force acts on the charged particle in the direction perpendicular to the direction of motion of the particle and this causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $90^{\circ}$. Then, $\quad \mathrm{W}=\mathrm{Fd} \cos 90^{\circ}$ $\mathrm{W}=0$ So, $\quad$ K.E. $=$ Work done $\text { K.E. }=0$ The change in kinetic energy of the charged particle after entering the magnetic field is zero. Hence, the kinetic energy of the charged particle after 3 second is $\mathrm{T}$.
COMEDK 2017
Moving Charges & Magnetism
153611
A magnetic field (B) perpendicular to plane of paper is present at a certain place. Calcium ion $\left(\mathrm{Ca}^{++}\right)$is moving in the plane with a velocity $v$. The magnetic force acting on it is
1 evB
2 $2 \mathrm{evB}$
3 $\frac{\text { evB }}{2}$
4 zero
Explanation:
B Magnetic force acting on a charge q moving with velocity $\overrightarrow{\mathrm{V}}$ in a magnetic field $\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \quad(\because \mathrm{a} \times \mathrm{b}=\mathrm{ab} \sin \theta)$ $\mathrm{F}=\mathrm{qvB} \sin \theta$ Where, $\mathrm{B}=$ flux density of magnetic field $\mathrm{q}=$ charge of the particle $\mathrm{v}=$ velocity of particle $\theta=$ angle between vector $\vec{v}$ and $\vec{B}$ For, charge on Calcium ion $\left(\mathrm{Ca}^{++}\right), \mathrm{q}=2 \mathrm{e}$ $\mathrm{F} =2 \mathrm{evB} \sin 90^{\circ}$ $\mathrm{F} =2 \mathrm{evB}$
COMEDK 2017
Moving Charges & Magnetism
153612
A charged particle moves through a magnetic field in a direction perpendicular to it. Then it
1 velocity remains unchanged
2 speed of the particle remains unchanged
3 direction of the particle remains unchanged
4 acceleration remains unchanged
Explanation:
B Given that, $\theta=90^{\circ}$ Force on a charged particle $\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{qvB} \sin \theta$ $\mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB}$ As force is acting perpendicular to velocity, this implies speed will not be changed only direction will be changed. Thus, speed of the particle remains unchanged.
153608
A proton and an $\alpha$-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. The ratio of the radii of their paths of proton to that of the $\alpha$-particle is
1 $2: 1$
2 $1: 2$
3 $1: 1$
4 $2: 3$
Explanation:
C From question, we know - Radius of circular path, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{Bq}}$ For same kinetic energy $\mathrm{K}$ and field $\mathrm{B}$, $r \propto \frac{\sqrt{m}}{q}$ $\text { So, } \quad \frac{r_{p}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{p}} \sqrt{\frac{m_{p}}{m_{\alpha}}}=\frac{2 e}{e} \sqrt{\frac{m_{p}}{4 m_{p}}}=\frac{1}{1}$ $r_{p}: r_{\alpha}=1: 1$
COMEDK 2014
Moving Charges & Magnetism
153609
What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, is required to make the electrons travel in a circular arc of radius $0.35 \mathrm{~m}$ ?
1 $2.1 \times 10^{-5} \mathrm{G}$
2 $6 \times 10^{-6} \mathrm{~T}$
3 $2.1 \times 10^{-5} \mathrm{~T}$
4 $6 \times 10^{-6} \mathrm{G}$
Explanation:
C Given that, $\mathrm{v}=1.3 \times 10^{6} \mathrm{~m} / \mathrm{sec}$ $\mathrm{r}=0.35 \mathrm{~m}$ Mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ Charge of electron $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ $\theta=90^{\circ}$ We know that, magnetic field in a circular arc, $\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qr}}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^{6}}{1.6 \times 10^{-19} \times 0.350}$ $\mathrm{~B}=\frac{11.83}{0.56} \times 10^{-6}$ $\mathrm{~B}=2.11 \times 10^{-5} \mathrm{~T}$
COMEDK 2016
Moving Charges & Magnetism
153610
A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be
1 $\mathrm{T}$
2 $4 \mathrm{~T}$
3 $3 \mathrm{~T}$
4 $2 \mathrm{~T}$
Explanation:
A When a charged particle enters in an uniform magnetic field, a magnetic force acts on the charged particle in the direction perpendicular to the direction of motion of the particle and this causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $90^{\circ}$. Then, $\quad \mathrm{W}=\mathrm{Fd} \cos 90^{\circ}$ $\mathrm{W}=0$ So, $\quad$ K.E. $=$ Work done $\text { K.E. }=0$ The change in kinetic energy of the charged particle after entering the magnetic field is zero. Hence, the kinetic energy of the charged particle after 3 second is $\mathrm{T}$.
COMEDK 2017
Moving Charges & Magnetism
153611
A magnetic field (B) perpendicular to plane of paper is present at a certain place. Calcium ion $\left(\mathrm{Ca}^{++}\right)$is moving in the plane with a velocity $v$. The magnetic force acting on it is
1 evB
2 $2 \mathrm{evB}$
3 $\frac{\text { evB }}{2}$
4 zero
Explanation:
B Magnetic force acting on a charge q moving with velocity $\overrightarrow{\mathrm{V}}$ in a magnetic field $\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \quad(\because \mathrm{a} \times \mathrm{b}=\mathrm{ab} \sin \theta)$ $\mathrm{F}=\mathrm{qvB} \sin \theta$ Where, $\mathrm{B}=$ flux density of magnetic field $\mathrm{q}=$ charge of the particle $\mathrm{v}=$ velocity of particle $\theta=$ angle between vector $\vec{v}$ and $\vec{B}$ For, charge on Calcium ion $\left(\mathrm{Ca}^{++}\right), \mathrm{q}=2 \mathrm{e}$ $\mathrm{F} =2 \mathrm{evB} \sin 90^{\circ}$ $\mathrm{F} =2 \mathrm{evB}$
COMEDK 2017
Moving Charges & Magnetism
153612
A charged particle moves through a magnetic field in a direction perpendicular to it. Then it
1 velocity remains unchanged
2 speed of the particle remains unchanged
3 direction of the particle remains unchanged
4 acceleration remains unchanged
Explanation:
B Given that, $\theta=90^{\circ}$ Force on a charged particle $\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{qvB} \sin \theta$ $\mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB}$ As force is acting perpendicular to velocity, this implies speed will not be changed only direction will be changed. Thus, speed of the particle remains unchanged.
153608
A proton and an $\alpha$-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. The ratio of the radii of their paths of proton to that of the $\alpha$-particle is
1 $2: 1$
2 $1: 2$
3 $1: 1$
4 $2: 3$
Explanation:
C From question, we know - Radius of circular path, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{Bq}}$ For same kinetic energy $\mathrm{K}$ and field $\mathrm{B}$, $r \propto \frac{\sqrt{m}}{q}$ $\text { So, } \quad \frac{r_{p}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{p}} \sqrt{\frac{m_{p}}{m_{\alpha}}}=\frac{2 e}{e} \sqrt{\frac{m_{p}}{4 m_{p}}}=\frac{1}{1}$ $r_{p}: r_{\alpha}=1: 1$
COMEDK 2014
Moving Charges & Magnetism
153609
What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, is required to make the electrons travel in a circular arc of radius $0.35 \mathrm{~m}$ ?
1 $2.1 \times 10^{-5} \mathrm{G}$
2 $6 \times 10^{-6} \mathrm{~T}$
3 $2.1 \times 10^{-5} \mathrm{~T}$
4 $6 \times 10^{-6} \mathrm{G}$
Explanation:
C Given that, $\mathrm{v}=1.3 \times 10^{6} \mathrm{~m} / \mathrm{sec}$ $\mathrm{r}=0.35 \mathrm{~m}$ Mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ Charge of electron $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ $\theta=90^{\circ}$ We know that, magnetic field in a circular arc, $\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qr}}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^{6}}{1.6 \times 10^{-19} \times 0.350}$ $\mathrm{~B}=\frac{11.83}{0.56} \times 10^{-6}$ $\mathrm{~B}=2.11 \times 10^{-5} \mathrm{~T}$
COMEDK 2016
Moving Charges & Magnetism
153610
A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be
1 $\mathrm{T}$
2 $4 \mathrm{~T}$
3 $3 \mathrm{~T}$
4 $2 \mathrm{~T}$
Explanation:
A When a charged particle enters in an uniform magnetic field, a magnetic force acts on the charged particle in the direction perpendicular to the direction of motion of the particle and this causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $90^{\circ}$. Then, $\quad \mathrm{W}=\mathrm{Fd} \cos 90^{\circ}$ $\mathrm{W}=0$ So, $\quad$ K.E. $=$ Work done $\text { K.E. }=0$ The change in kinetic energy of the charged particle after entering the magnetic field is zero. Hence, the kinetic energy of the charged particle after 3 second is $\mathrm{T}$.
COMEDK 2017
Moving Charges & Magnetism
153611
A magnetic field (B) perpendicular to plane of paper is present at a certain place. Calcium ion $\left(\mathrm{Ca}^{++}\right)$is moving in the plane with a velocity $v$. The magnetic force acting on it is
1 evB
2 $2 \mathrm{evB}$
3 $\frac{\text { evB }}{2}$
4 zero
Explanation:
B Magnetic force acting on a charge q moving with velocity $\overrightarrow{\mathrm{V}}$ in a magnetic field $\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \quad(\because \mathrm{a} \times \mathrm{b}=\mathrm{ab} \sin \theta)$ $\mathrm{F}=\mathrm{qvB} \sin \theta$ Where, $\mathrm{B}=$ flux density of magnetic field $\mathrm{q}=$ charge of the particle $\mathrm{v}=$ velocity of particle $\theta=$ angle between vector $\vec{v}$ and $\vec{B}$ For, charge on Calcium ion $\left(\mathrm{Ca}^{++}\right), \mathrm{q}=2 \mathrm{e}$ $\mathrm{F} =2 \mathrm{evB} \sin 90^{\circ}$ $\mathrm{F} =2 \mathrm{evB}$
COMEDK 2017
Moving Charges & Magnetism
153612
A charged particle moves through a magnetic field in a direction perpendicular to it. Then it
1 velocity remains unchanged
2 speed of the particle remains unchanged
3 direction of the particle remains unchanged
4 acceleration remains unchanged
Explanation:
B Given that, $\theta=90^{\circ}$ Force on a charged particle $\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{qvB} \sin \theta$ $\mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB}$ As force is acting perpendicular to velocity, this implies speed will not be changed only direction will be changed. Thus, speed of the particle remains unchanged.
153608
A proton and an $\alpha$-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. The ratio of the radii of their paths of proton to that of the $\alpha$-particle is
1 $2: 1$
2 $1: 2$
3 $1: 1$
4 $2: 3$
Explanation:
C From question, we know - Radius of circular path, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}{\mathrm{Bq}}$ For same kinetic energy $\mathrm{K}$ and field $\mathrm{B}$, $r \propto \frac{\sqrt{m}}{q}$ $\text { So, } \quad \frac{r_{p}}{r_{\alpha}}=\frac{q_{\alpha}}{q_{p}} \sqrt{\frac{m_{p}}{m_{\alpha}}}=\frac{2 e}{e} \sqrt{\frac{m_{p}}{4 m_{p}}}=\frac{1}{1}$ $r_{p}: r_{\alpha}=1: 1$
COMEDK 2014
Moving Charges & Magnetism
153609
What uniform magnetic field applied perpendicular to a beam of electrons moving at $1.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$, is required to make the electrons travel in a circular arc of radius $0.35 \mathrm{~m}$ ?
1 $2.1 \times 10^{-5} \mathrm{G}$
2 $6 \times 10^{-6} \mathrm{~T}$
3 $2.1 \times 10^{-5} \mathrm{~T}$
4 $6 \times 10^{-6} \mathrm{G}$
Explanation:
C Given that, $\mathrm{v}=1.3 \times 10^{6} \mathrm{~m} / \mathrm{sec}$ $\mathrm{r}=0.35 \mathrm{~m}$ Mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ Charge of electron $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ $\theta=90^{\circ}$ We know that, magnetic field in a circular arc, $\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qr}}=\frac{9.1 \times 10^{-31} \times 1.3 \times 10^{6}}{1.6 \times 10^{-19} \times 0.350}$ $\mathrm{~B}=\frac{11.83}{0.56} \times 10^{-6}$ $\mathrm{~B}=2.11 \times 10^{-5} \mathrm{~T}$
COMEDK 2016
Moving Charges & Magnetism
153610
A particle of mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be
1 $\mathrm{T}$
2 $4 \mathrm{~T}$
3 $3 \mathrm{~T}$
4 $2 \mathrm{~T}$
Explanation:
A When a charged particle enters in an uniform magnetic field, a magnetic force acts on the charged particle in the direction perpendicular to the direction of motion of the particle and this causes the charged particle to perform uniform circular motion in the magnetic field. The angle between the magnetic force and the displacement of the particle is $90^{\circ}$. Then, $\quad \mathrm{W}=\mathrm{Fd} \cos 90^{\circ}$ $\mathrm{W}=0$ So, $\quad$ K.E. $=$ Work done $\text { K.E. }=0$ The change in kinetic energy of the charged particle after entering the magnetic field is zero. Hence, the kinetic energy of the charged particle after 3 second is $\mathrm{T}$.
COMEDK 2017
Moving Charges & Magnetism
153611
A magnetic field (B) perpendicular to plane of paper is present at a certain place. Calcium ion $\left(\mathrm{Ca}^{++}\right)$is moving in the plane with a velocity $v$. The magnetic force acting on it is
1 evB
2 $2 \mathrm{evB}$
3 $\frac{\text { evB }}{2}$
4 zero
Explanation:
B Magnetic force acting on a charge q moving with velocity $\overrightarrow{\mathrm{V}}$ in a magnetic field $\overrightarrow{\mathrm{B}}$, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \quad(\because \mathrm{a} \times \mathrm{b}=\mathrm{ab} \sin \theta)$ $\mathrm{F}=\mathrm{qvB} \sin \theta$ Where, $\mathrm{B}=$ flux density of magnetic field $\mathrm{q}=$ charge of the particle $\mathrm{v}=$ velocity of particle $\theta=$ angle between vector $\vec{v}$ and $\vec{B}$ For, charge on Calcium ion $\left(\mathrm{Ca}^{++}\right), \mathrm{q}=2 \mathrm{e}$ $\mathrm{F} =2 \mathrm{evB} \sin 90^{\circ}$ $\mathrm{F} =2 \mathrm{evB}$
COMEDK 2017
Moving Charges & Magnetism
153612
A charged particle moves through a magnetic field in a direction perpendicular to it. Then it
1 velocity remains unchanged
2 speed of the particle remains unchanged
3 direction of the particle remains unchanged
4 acceleration remains unchanged
Explanation:
B Given that, $\theta=90^{\circ}$ Force on a charged particle $\mathrm{F}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{qvB} \sin \theta$ $\mathrm{F}=\mathrm{qvB} \sin 90^{\circ}=\mathrm{qvB}$ As force is acting perpendicular to velocity, this implies speed will not be changed only direction will be changed. Thus, speed of the particle remains unchanged.
