153528
A proton, a deuteron and $\alpha$ particle accelerated through the same potential difference enter the region of uniform magnetic field, moving at right angles to magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 1$
2 $1: 1: 2$
3 $2: 1: 2$
4 $2: 2: 1$
Explanation:
B We know that, Charge on proton $\left(\mathrm{q}_{\mathrm{p}}\right)=+\mathrm{e}$ Charge on deuteron $\left(\mathrm{q}_{\mathrm{d}}\right)=+\mathrm{e}$ Charge on $\alpha$ particle $\left(\mathrm{q}_{\alpha}\right)=+2 \mathrm{e}$ $\therefore \quad \mathrm{K}=\mathrm{q} \Delta \mathrm{V}$ According to the question potential difference is same So, $\quad \mathrm{K} \propto \mathrm{q}$ $\mathrm{K}_{\mathrm{P}}: \mathrm{K}_{\mathrm{d}}: \mathrm{K}_{\alpha}=+\mathrm{e}:+\mathrm{e}:+2 \mathrm{e}$ $=1: 1: 2$
MHT-CET 2020
Moving Charges & Magnetism
153529
An electron of charge ' $e$ ' and mass ' $m$ ' is accelerated through a potential difference of ' $V$ ' volt. It enters a transverse uniform magnetic field ' $B$ ' and describes a circular path. The radius of the circular path is
D Kinetic energy will be equal to $\mathrm{eV}$ $\because \quad \frac{1}{2} \mathrm{mV}^{2}=\mathrm{eV}$ Or $\quad \mathrm{V}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ And $\quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$ Putting the value of ' $\mathrm{V}$ ' we get - $\mathrm{R}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{eB}^{2}}}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{e}}}$
MHT-CET 2020
Moving Charges & Magnetism
153530
An electron moves with a speed of $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in a magnetic field $\vec{B}=(\hat{i}-4 \hat{j}-3 \hat{k}) T$. The magnitude of the force (in Newton) experienced by the electron is
1 $1.18 \times 10^{-13}$
2 $1.28 \times 10^{-13}$
3 $1.6 \times 10^{-13}$
4 $1.72 \times 10^{-13}$
Explanation:
C Given, $\overrightarrow{\mathrm{v}}=2 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{sec}$ $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{T}$ Force on charge particle in magnetic field is- $\vec{F}=q(\vec{v} \times \vec{B})$ $=1.6 \times 10^{-19}[\vec{v} \hat{i} \times(\hat{i}-4 \hat{j}-3 \hat{k})]$ $=16 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times v$ $=1.6 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times 2 \times 10^{5}$ $|F|=1.6 \times 10^{-13} \mathrm{~N}$
COMEDK 2020
Moving Charges & Magnetism
153532
An electron of mass $m$ is accelerated by a d.c. potential difference $V$ and then subjected to a transverse magnetic field $B$, The trajectory of the electron will be a circle of radius
D Given, mass $=\mathrm{m}$, potential difference $=\mathrm{V}$, magnetic field $=\mathrm{B}$ We know that, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$ and K.E. $=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Put the value of $v$ in equation (i), $\mathrm{r}=\frac{\mathrm{m} \times \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{qB}}$ $\mathrm{r}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^{2}}} \quad(\because \mathrm{q}=\mathrm{e})$
153528
A proton, a deuteron and $\alpha$ particle accelerated through the same potential difference enter the region of uniform magnetic field, moving at right angles to magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 1$
2 $1: 1: 2$
3 $2: 1: 2$
4 $2: 2: 1$
Explanation:
B We know that, Charge on proton $\left(\mathrm{q}_{\mathrm{p}}\right)=+\mathrm{e}$ Charge on deuteron $\left(\mathrm{q}_{\mathrm{d}}\right)=+\mathrm{e}$ Charge on $\alpha$ particle $\left(\mathrm{q}_{\alpha}\right)=+2 \mathrm{e}$ $\therefore \quad \mathrm{K}=\mathrm{q} \Delta \mathrm{V}$ According to the question potential difference is same So, $\quad \mathrm{K} \propto \mathrm{q}$ $\mathrm{K}_{\mathrm{P}}: \mathrm{K}_{\mathrm{d}}: \mathrm{K}_{\alpha}=+\mathrm{e}:+\mathrm{e}:+2 \mathrm{e}$ $=1: 1: 2$
MHT-CET 2020
Moving Charges & Magnetism
153529
An electron of charge ' $e$ ' and mass ' $m$ ' is accelerated through a potential difference of ' $V$ ' volt. It enters a transverse uniform magnetic field ' $B$ ' and describes a circular path. The radius of the circular path is
D Kinetic energy will be equal to $\mathrm{eV}$ $\because \quad \frac{1}{2} \mathrm{mV}^{2}=\mathrm{eV}$ Or $\quad \mathrm{V}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ And $\quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$ Putting the value of ' $\mathrm{V}$ ' we get - $\mathrm{R}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{eB}^{2}}}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{e}}}$
MHT-CET 2020
Moving Charges & Magnetism
153530
An electron moves with a speed of $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in a magnetic field $\vec{B}=(\hat{i}-4 \hat{j}-3 \hat{k}) T$. The magnitude of the force (in Newton) experienced by the electron is
1 $1.18 \times 10^{-13}$
2 $1.28 \times 10^{-13}$
3 $1.6 \times 10^{-13}$
4 $1.72 \times 10^{-13}$
Explanation:
C Given, $\overrightarrow{\mathrm{v}}=2 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{sec}$ $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{T}$ Force on charge particle in magnetic field is- $\vec{F}=q(\vec{v} \times \vec{B})$ $=1.6 \times 10^{-19}[\vec{v} \hat{i} \times(\hat{i}-4 \hat{j}-3 \hat{k})]$ $=16 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times v$ $=1.6 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times 2 \times 10^{5}$ $|F|=1.6 \times 10^{-13} \mathrm{~N}$
COMEDK 2020
Moving Charges & Magnetism
153532
An electron of mass $m$ is accelerated by a d.c. potential difference $V$ and then subjected to a transverse magnetic field $B$, The trajectory of the electron will be a circle of radius
D Given, mass $=\mathrm{m}$, potential difference $=\mathrm{V}$, magnetic field $=\mathrm{B}$ We know that, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$ and K.E. $=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Put the value of $v$ in equation (i), $\mathrm{r}=\frac{\mathrm{m} \times \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{qB}}$ $\mathrm{r}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^{2}}} \quad(\because \mathrm{q}=\mathrm{e})$
153528
A proton, a deuteron and $\alpha$ particle accelerated through the same potential difference enter the region of uniform magnetic field, moving at right angles to magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 1$
2 $1: 1: 2$
3 $2: 1: 2$
4 $2: 2: 1$
Explanation:
B We know that, Charge on proton $\left(\mathrm{q}_{\mathrm{p}}\right)=+\mathrm{e}$ Charge on deuteron $\left(\mathrm{q}_{\mathrm{d}}\right)=+\mathrm{e}$ Charge on $\alpha$ particle $\left(\mathrm{q}_{\alpha}\right)=+2 \mathrm{e}$ $\therefore \quad \mathrm{K}=\mathrm{q} \Delta \mathrm{V}$ According to the question potential difference is same So, $\quad \mathrm{K} \propto \mathrm{q}$ $\mathrm{K}_{\mathrm{P}}: \mathrm{K}_{\mathrm{d}}: \mathrm{K}_{\alpha}=+\mathrm{e}:+\mathrm{e}:+2 \mathrm{e}$ $=1: 1: 2$
MHT-CET 2020
Moving Charges & Magnetism
153529
An electron of charge ' $e$ ' and mass ' $m$ ' is accelerated through a potential difference of ' $V$ ' volt. It enters a transverse uniform magnetic field ' $B$ ' and describes a circular path. The radius of the circular path is
D Kinetic energy will be equal to $\mathrm{eV}$ $\because \quad \frac{1}{2} \mathrm{mV}^{2}=\mathrm{eV}$ Or $\quad \mathrm{V}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ And $\quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$ Putting the value of ' $\mathrm{V}$ ' we get - $\mathrm{R}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{eB}^{2}}}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{e}}}$
MHT-CET 2020
Moving Charges & Magnetism
153530
An electron moves with a speed of $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in a magnetic field $\vec{B}=(\hat{i}-4 \hat{j}-3 \hat{k}) T$. The magnitude of the force (in Newton) experienced by the electron is
1 $1.18 \times 10^{-13}$
2 $1.28 \times 10^{-13}$
3 $1.6 \times 10^{-13}$
4 $1.72 \times 10^{-13}$
Explanation:
C Given, $\overrightarrow{\mathrm{v}}=2 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{sec}$ $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{T}$ Force on charge particle in magnetic field is- $\vec{F}=q(\vec{v} \times \vec{B})$ $=1.6 \times 10^{-19}[\vec{v} \hat{i} \times(\hat{i}-4 \hat{j}-3 \hat{k})]$ $=16 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times v$ $=1.6 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times 2 \times 10^{5}$ $|F|=1.6 \times 10^{-13} \mathrm{~N}$
COMEDK 2020
Moving Charges & Magnetism
153532
An electron of mass $m$ is accelerated by a d.c. potential difference $V$ and then subjected to a transverse magnetic field $B$, The trajectory of the electron will be a circle of radius
D Given, mass $=\mathrm{m}$, potential difference $=\mathrm{V}$, magnetic field $=\mathrm{B}$ We know that, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$ and K.E. $=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Put the value of $v$ in equation (i), $\mathrm{r}=\frac{\mathrm{m} \times \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{qB}}$ $\mathrm{r}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^{2}}} \quad(\because \mathrm{q}=\mathrm{e})$
153528
A proton, a deuteron and $\alpha$ particle accelerated through the same potential difference enter the region of uniform magnetic field, moving at right angles to magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 1$
2 $1: 1: 2$
3 $2: 1: 2$
4 $2: 2: 1$
Explanation:
B We know that, Charge on proton $\left(\mathrm{q}_{\mathrm{p}}\right)=+\mathrm{e}$ Charge on deuteron $\left(\mathrm{q}_{\mathrm{d}}\right)=+\mathrm{e}$ Charge on $\alpha$ particle $\left(\mathrm{q}_{\alpha}\right)=+2 \mathrm{e}$ $\therefore \quad \mathrm{K}=\mathrm{q} \Delta \mathrm{V}$ According to the question potential difference is same So, $\quad \mathrm{K} \propto \mathrm{q}$ $\mathrm{K}_{\mathrm{P}}: \mathrm{K}_{\mathrm{d}}: \mathrm{K}_{\alpha}=+\mathrm{e}:+\mathrm{e}:+2 \mathrm{e}$ $=1: 1: 2$
MHT-CET 2020
Moving Charges & Magnetism
153529
An electron of charge ' $e$ ' and mass ' $m$ ' is accelerated through a potential difference of ' $V$ ' volt. It enters a transverse uniform magnetic field ' $B$ ' and describes a circular path. The radius of the circular path is
D Kinetic energy will be equal to $\mathrm{eV}$ $\because \quad \frac{1}{2} \mathrm{mV}^{2}=\mathrm{eV}$ Or $\quad \mathrm{V}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ And $\quad \mathrm{R}=\frac{\mathrm{mV}}{\mathrm{eB}}$ Putting the value of ' $\mathrm{V}$ ' we get - $\mathrm{R}=\sqrt{\frac{2 \mathrm{mV}}{\mathrm{eB}^{2}}}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{e}}}$
MHT-CET 2020
Moving Charges & Magnetism
153530
An electron moves with a speed of $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ along the positive $x$-direction in a magnetic field $\vec{B}=(\hat{i}-4 \hat{j}-3 \hat{k}) T$. The magnitude of the force (in Newton) experienced by the electron is
1 $1.18 \times 10^{-13}$
2 $1.28 \times 10^{-13}$
3 $1.6 \times 10^{-13}$
4 $1.72 \times 10^{-13}$
Explanation:
C Given, $\overrightarrow{\mathrm{v}}=2 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{sec}$ $\overrightarrow{\mathrm{B}}=(\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \mathrm{T}$ Force on charge particle in magnetic field is- $\vec{F}=q(\vec{v} \times \vec{B})$ $=1.6 \times 10^{-19}[\vec{v} \hat{i} \times(\hat{i}-4 \hat{j}-3 \hat{k})]$ $=16 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times v$ $=1.6 \times 10^{-19}(-4 \hat{k}+3 \hat{j}) \times 2 \times 10^{5}$ $|F|=1.6 \times 10^{-13} \mathrm{~N}$
COMEDK 2020
Moving Charges & Magnetism
153532
An electron of mass $m$ is accelerated by a d.c. potential difference $V$ and then subjected to a transverse magnetic field $B$, The trajectory of the electron will be a circle of radius
D Given, mass $=\mathrm{m}$, potential difference $=\mathrm{V}$, magnetic field $=\mathrm{B}$ We know that, $\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$ and K.E. $=\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Put the value of $v$ in equation (i), $\mathrm{r}=\frac{\mathrm{m} \times \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}}{\mathrm{qB}}$ $\mathrm{r}=\sqrt{\frac{2 \mathrm{Vm}}{\mathrm{eB}^{2}}} \quad(\because \mathrm{q}=\mathrm{e})$
