153718
The intensity of magnetic field due to an isolated pole of strength $m$ at a point distant $r$ from it will be
1 $\mathrm{m} / \mathrm{r}^{2}$
2 $\mathrm{mr}^{2}$
3 $\mathrm{r}^{2} / \mathrm{m}$
4 $\mathrm{m} / \mathrm{r}$
Explanation:
A Given, pole of strength $=\mathrm{m}$ We know that, The intensity of magnetic field due to a single pole is given by $\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$
J and K CET- 2005
Moving Charges & Magnetism
153719
A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 3$
2 $1: 1: 2$
3 $1: 1: 1$
4 $1: 2: 4$
Explanation:
B Kinetic energy gained by charged particle of charge $\mathrm{q}$ when accelerated under a potential difference $\mathrm{v}$ will be K.E. = qv For given velocity $\mathrm{KE} \propto \mathrm{q}$ For proton, deuteron and $\alpha$ - particle the Ratio of charge, ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{2} \mathrm{He}^{4}$ $\mathrm{q}_{\mathrm{p}}: \mathrm{q}_{\mathrm{d}}: \mathrm{q}_{\alpha}=1: 1: 2$
J and K CET- 2005
Moving Charges & Magnetism
153723
A charged particle moves along a circle under the action of magnetic and electric fields, then this region of space may have
1 $\mathrm{E}=0, \mathrm{~B}=0$
2 $\mathrm{E}=0, \mathrm{~B} \neq 0$
3 $\mathrm{E} \neq 0, \mathrm{~B}=0$
4 $\mathrm{E} \neq 0, \mathrm{~B} \neq 0$
Explanation:
B A charged particle moves along a circle. That means magnetic force provides centripetal force and due to centripetal force particle moves in circular path. The particle can never move in circular path due to electric field because the radius of circular orbit will increasing due to electric force. So, the particle moves in circular part when, $\mathrm{E}=0, \mathrm{~B} \neq 0$
UP CPMT-2010
Moving Charges & Magnetism
153729
A charge $q$ moves with velocity $\vec{v}$ in the direction of the applied field $\vec{B}$. The force that acts on the charge is
D We know that, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ $F=q v B \sin \theta$ $\theta=0^{\circ}$ $F=0$ The force act on the charge is zero.
153718
The intensity of magnetic field due to an isolated pole of strength $m$ at a point distant $r$ from it will be
1 $\mathrm{m} / \mathrm{r}^{2}$
2 $\mathrm{mr}^{2}$
3 $\mathrm{r}^{2} / \mathrm{m}$
4 $\mathrm{m} / \mathrm{r}$
Explanation:
A Given, pole of strength $=\mathrm{m}$ We know that, The intensity of magnetic field due to a single pole is given by $\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$
J and K CET- 2005
Moving Charges & Magnetism
153719
A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 3$
2 $1: 1: 2$
3 $1: 1: 1$
4 $1: 2: 4$
Explanation:
B Kinetic energy gained by charged particle of charge $\mathrm{q}$ when accelerated under a potential difference $\mathrm{v}$ will be K.E. = qv For given velocity $\mathrm{KE} \propto \mathrm{q}$ For proton, deuteron and $\alpha$ - particle the Ratio of charge, ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{2} \mathrm{He}^{4}$ $\mathrm{q}_{\mathrm{p}}: \mathrm{q}_{\mathrm{d}}: \mathrm{q}_{\alpha}=1: 1: 2$
J and K CET- 2005
Moving Charges & Magnetism
153723
A charged particle moves along a circle under the action of magnetic and electric fields, then this region of space may have
1 $\mathrm{E}=0, \mathrm{~B}=0$
2 $\mathrm{E}=0, \mathrm{~B} \neq 0$
3 $\mathrm{E} \neq 0, \mathrm{~B}=0$
4 $\mathrm{E} \neq 0, \mathrm{~B} \neq 0$
Explanation:
B A charged particle moves along a circle. That means magnetic force provides centripetal force and due to centripetal force particle moves in circular path. The particle can never move in circular path due to electric field because the radius of circular orbit will increasing due to electric force. So, the particle moves in circular part when, $\mathrm{E}=0, \mathrm{~B} \neq 0$
UP CPMT-2010
Moving Charges & Magnetism
153729
A charge $q$ moves with velocity $\vec{v}$ in the direction of the applied field $\vec{B}$. The force that acts on the charge is
D We know that, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ $F=q v B \sin \theta$ $\theta=0^{\circ}$ $F=0$ The force act on the charge is zero.
153718
The intensity of magnetic field due to an isolated pole of strength $m$ at a point distant $r$ from it will be
1 $\mathrm{m} / \mathrm{r}^{2}$
2 $\mathrm{mr}^{2}$
3 $\mathrm{r}^{2} / \mathrm{m}$
4 $\mathrm{m} / \mathrm{r}$
Explanation:
A Given, pole of strength $=\mathrm{m}$ We know that, The intensity of magnetic field due to a single pole is given by $\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$
J and K CET- 2005
Moving Charges & Magnetism
153719
A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 3$
2 $1: 1: 2$
3 $1: 1: 1$
4 $1: 2: 4$
Explanation:
B Kinetic energy gained by charged particle of charge $\mathrm{q}$ when accelerated under a potential difference $\mathrm{v}$ will be K.E. = qv For given velocity $\mathrm{KE} \propto \mathrm{q}$ For proton, deuteron and $\alpha$ - particle the Ratio of charge, ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{2} \mathrm{He}^{4}$ $\mathrm{q}_{\mathrm{p}}: \mathrm{q}_{\mathrm{d}}: \mathrm{q}_{\alpha}=1: 1: 2$
J and K CET- 2005
Moving Charges & Magnetism
153723
A charged particle moves along a circle under the action of magnetic and electric fields, then this region of space may have
1 $\mathrm{E}=0, \mathrm{~B}=0$
2 $\mathrm{E}=0, \mathrm{~B} \neq 0$
3 $\mathrm{E} \neq 0, \mathrm{~B}=0$
4 $\mathrm{E} \neq 0, \mathrm{~B} \neq 0$
Explanation:
B A charged particle moves along a circle. That means magnetic force provides centripetal force and due to centripetal force particle moves in circular path. The particle can never move in circular path due to electric field because the radius of circular orbit will increasing due to electric force. So, the particle moves in circular part when, $\mathrm{E}=0, \mathrm{~B} \neq 0$
UP CPMT-2010
Moving Charges & Magnetism
153729
A charge $q$ moves with velocity $\vec{v}$ in the direction of the applied field $\vec{B}$. The force that acts on the charge is
D We know that, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ $F=q v B \sin \theta$ $\theta=0^{\circ}$ $F=0$ The force act on the charge is zero.
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Moving Charges & Magnetism
153718
The intensity of magnetic field due to an isolated pole of strength $m$ at a point distant $r$ from it will be
1 $\mathrm{m} / \mathrm{r}^{2}$
2 $\mathrm{mr}^{2}$
3 $\mathrm{r}^{2} / \mathrm{m}$
4 $\mathrm{m} / \mathrm{r}$
Explanation:
A Given, pole of strength $=\mathrm{m}$ We know that, The intensity of magnetic field due to a single pole is given by $\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{m}}{\mathrm{r}^{2}}$ $\mathrm{~B} \propto \frac{\mathrm{m}}{\mathrm{r}^{2}}$
J and K CET- 2005
Moving Charges & Magnetism
153719
A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter in a normal uniform magnetic field. The ratio of their kinetic energies will be
1 $2: 1: 3$
2 $1: 1: 2$
3 $1: 1: 1$
4 $1: 2: 4$
Explanation:
B Kinetic energy gained by charged particle of charge $\mathrm{q}$ when accelerated under a potential difference $\mathrm{v}$ will be K.E. = qv For given velocity $\mathrm{KE} \propto \mathrm{q}$ For proton, deuteron and $\alpha$ - particle the Ratio of charge, ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{2} \mathrm{He}^{4}$ $\mathrm{q}_{\mathrm{p}}: \mathrm{q}_{\mathrm{d}}: \mathrm{q}_{\alpha}=1: 1: 2$
J and K CET- 2005
Moving Charges & Magnetism
153723
A charged particle moves along a circle under the action of magnetic and electric fields, then this region of space may have
1 $\mathrm{E}=0, \mathrm{~B}=0$
2 $\mathrm{E}=0, \mathrm{~B} \neq 0$
3 $\mathrm{E} \neq 0, \mathrm{~B}=0$
4 $\mathrm{E} \neq 0, \mathrm{~B} \neq 0$
Explanation:
B A charged particle moves along a circle. That means magnetic force provides centripetal force and due to centripetal force particle moves in circular path. The particle can never move in circular path due to electric field because the radius of circular orbit will increasing due to electric force. So, the particle moves in circular part when, $\mathrm{E}=0, \mathrm{~B} \neq 0$
UP CPMT-2010
Moving Charges & Magnetism
153729
A charge $q$ moves with velocity $\vec{v}$ in the direction of the applied field $\vec{B}$. The force that acts on the charge is
D We know that, $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ $F=q v B \sin \theta$ $\theta=0^{\circ}$ $F=0$ The force act on the charge is zero.
