NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153553
Assertion: Magnetic field cannot change K.E. of moving charge. Reason: Magnetic field cannot change velocity vector.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
C - Magnetic field cannot change K.E. of moving charge. - Magnetic field can change the direction of velocity vector while magnitude remains same. Hence, option (c) is correct.
AIIMS-26.05.2018(M)
Moving Charges & Magnetism
153554
A beam of protons with velocity $600 \mathrm{~km} / \mathrm{s}$ enters a uniform magnetic field of $0.2 \mathrm{~T}$. the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be
153555
An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$. Suddenly, the field is reduced to $\frac{B}{2}$. The radius of the circular path now becomes :
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $\frac{r}{4}$
4 $4 \mathrm{r}$
Explanation:
B Given, radius $=r$, magnetic field $=\mathrm{B}$ We know- The radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{Bq}}$ Here, $\mathrm{r} \propto \frac{\mathrm{mv}}{\mathrm{B}}$ Now magnetic field is reduced to $\frac{B}{2}$. So, new magnetic field $\left(\mathrm{B}^{\prime}\right)=\frac{\mathrm{B}}{2}$. Now, Radius of new path $\left(\mathrm{r}^{\prime}\right)=\frac{\mathrm{mv}}{\mathrm{Bq}}$ $r^{\prime} =\frac{m v}{\frac{B}{2} q}, \quad r^{\prime}=\frac{2 m v}{B q}$ $r^{\prime} =2 r$ Hence, new radius of circular path is $2 r$.
Karnataka CET-2018
Moving Charges & Magnetism
153556
A charge $q$ is accelerated through a potential difference $V$. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$. The potential difference required to move it in a circle of radius $2 r$ is :
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $3 \mathrm{~V}$
Explanation:
B Given, $\mathrm{r}_{1}=2 \mathrm{r}$ Radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$ $\mathrm{r} \propto \sqrt{\mathrm{V}} {[\because \mathrm{B}=\text { Constant }]}$ $\text { or } \mathrm{r}^{2} \propto \mathrm{V}$ $\text { So, } \frac{\mathrm{V}_{1}}{\mathrm{~V}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}}\right)^{2}=\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}=4$ $\text { or } \mathrm{V}_{1}=4 \mathrm{~V}$
153553
Assertion: Magnetic field cannot change K.E. of moving charge. Reason: Magnetic field cannot change velocity vector.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
C - Magnetic field cannot change K.E. of moving charge. - Magnetic field can change the direction of velocity vector while magnitude remains same. Hence, option (c) is correct.
AIIMS-26.05.2018(M)
Moving Charges & Magnetism
153554
A beam of protons with velocity $600 \mathrm{~km} / \mathrm{s}$ enters a uniform magnetic field of $0.2 \mathrm{~T}$. the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be
153555
An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$. Suddenly, the field is reduced to $\frac{B}{2}$. The radius of the circular path now becomes :
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $\frac{r}{4}$
4 $4 \mathrm{r}$
Explanation:
B Given, radius $=r$, magnetic field $=\mathrm{B}$ We know- The radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{Bq}}$ Here, $\mathrm{r} \propto \frac{\mathrm{mv}}{\mathrm{B}}$ Now magnetic field is reduced to $\frac{B}{2}$. So, new magnetic field $\left(\mathrm{B}^{\prime}\right)=\frac{\mathrm{B}}{2}$. Now, Radius of new path $\left(\mathrm{r}^{\prime}\right)=\frac{\mathrm{mv}}{\mathrm{Bq}}$ $r^{\prime} =\frac{m v}{\frac{B}{2} q}, \quad r^{\prime}=\frac{2 m v}{B q}$ $r^{\prime} =2 r$ Hence, new radius of circular path is $2 r$.
Karnataka CET-2018
Moving Charges & Magnetism
153556
A charge $q$ is accelerated through a potential difference $V$. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$. The potential difference required to move it in a circle of radius $2 r$ is :
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $3 \mathrm{~V}$
Explanation:
B Given, $\mathrm{r}_{1}=2 \mathrm{r}$ Radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$ $\mathrm{r} \propto \sqrt{\mathrm{V}} {[\because \mathrm{B}=\text { Constant }]}$ $\text { or } \mathrm{r}^{2} \propto \mathrm{V}$ $\text { So, } \frac{\mathrm{V}_{1}}{\mathrm{~V}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}}\right)^{2}=\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}=4$ $\text { or } \mathrm{V}_{1}=4 \mathrm{~V}$
153553
Assertion: Magnetic field cannot change K.E. of moving charge. Reason: Magnetic field cannot change velocity vector.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
C - Magnetic field cannot change K.E. of moving charge. - Magnetic field can change the direction of velocity vector while magnitude remains same. Hence, option (c) is correct.
AIIMS-26.05.2018(M)
Moving Charges & Magnetism
153554
A beam of protons with velocity $600 \mathrm{~km} / \mathrm{s}$ enters a uniform magnetic field of $0.2 \mathrm{~T}$. the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be
153555
An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$. Suddenly, the field is reduced to $\frac{B}{2}$. The radius of the circular path now becomes :
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $\frac{r}{4}$
4 $4 \mathrm{r}$
Explanation:
B Given, radius $=r$, magnetic field $=\mathrm{B}$ We know- The radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{Bq}}$ Here, $\mathrm{r} \propto \frac{\mathrm{mv}}{\mathrm{B}}$ Now magnetic field is reduced to $\frac{B}{2}$. So, new magnetic field $\left(\mathrm{B}^{\prime}\right)=\frac{\mathrm{B}}{2}$. Now, Radius of new path $\left(\mathrm{r}^{\prime}\right)=\frac{\mathrm{mv}}{\mathrm{Bq}}$ $r^{\prime} =\frac{m v}{\frac{B}{2} q}, \quad r^{\prime}=\frac{2 m v}{B q}$ $r^{\prime} =2 r$ Hence, new radius of circular path is $2 r$.
Karnataka CET-2018
Moving Charges & Magnetism
153556
A charge $q$ is accelerated through a potential difference $V$. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$. The potential difference required to move it in a circle of radius $2 r$ is :
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $3 \mathrm{~V}$
Explanation:
B Given, $\mathrm{r}_{1}=2 \mathrm{r}$ Radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$ $\mathrm{r} \propto \sqrt{\mathrm{V}} {[\because \mathrm{B}=\text { Constant }]}$ $\text { or } \mathrm{r}^{2} \propto \mathrm{V}$ $\text { So, } \frac{\mathrm{V}_{1}}{\mathrm{~V}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}}\right)^{2}=\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}=4$ $\text { or } \mathrm{V}_{1}=4 \mathrm{~V}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153553
Assertion: Magnetic field cannot change K.E. of moving charge. Reason: Magnetic field cannot change velocity vector.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect.
Explanation:
C - Magnetic field cannot change K.E. of moving charge. - Magnetic field can change the direction of velocity vector while magnitude remains same. Hence, option (c) is correct.
AIIMS-26.05.2018(M)
Moving Charges & Magnetism
153554
A beam of protons with velocity $600 \mathrm{~km} / \mathrm{s}$ enters a uniform magnetic field of $0.2 \mathrm{~T}$. the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be
153555
An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$. Suddenly, the field is reduced to $\frac{B}{2}$. The radius of the circular path now becomes :
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $\frac{r}{4}$
4 $4 \mathrm{r}$
Explanation:
B Given, radius $=r$, magnetic field $=\mathrm{B}$ We know- The radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{Bq}}$ Here, $\mathrm{r} \propto \frac{\mathrm{mv}}{\mathrm{B}}$ Now magnetic field is reduced to $\frac{B}{2}$. So, new magnetic field $\left(\mathrm{B}^{\prime}\right)=\frac{\mathrm{B}}{2}$. Now, Radius of new path $\left(\mathrm{r}^{\prime}\right)=\frac{\mathrm{mv}}{\mathrm{Bq}}$ $r^{\prime} =\frac{m v}{\frac{B}{2} q}, \quad r^{\prime}=\frac{2 m v}{B q}$ $r^{\prime} =2 r$ Hence, new radius of circular path is $2 r$.
Karnataka CET-2018
Moving Charges & Magnetism
153556
A charge $q$ is accelerated through a potential difference $V$. It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$. The potential difference required to move it in a circle of radius $2 r$ is :
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $1 \mathrm{~V}$
4 $3 \mathrm{~V}$
Explanation:
B Given, $\mathrm{r}_{1}=2 \mathrm{r}$ Radius of circular path $(\mathrm{r})=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}}$ $\mathrm{r} \propto \sqrt{\mathrm{V}} {[\because \mathrm{B}=\text { Constant }]}$ $\text { or } \mathrm{r}^{2} \propto \mathrm{V}$ $\text { So, } \frac{\mathrm{V}_{1}}{\mathrm{~V}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}}\right)^{2}=\left(\frac{2 \mathrm{r}}{\mathrm{r}}\right)^{2}=4$ $\text { or } \mathrm{V}_{1}=4 \mathrm{~V}$