NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153873
If a magnetic dipole of dipole moment $M$ rotated through an angle $\theta$ with respect to the direction of the field $H$, then the work done is :
1 $\mathrm{MH} \sin \theta$
2 $\mathrm{MH}(1-\sin \theta)$
3 $\mathrm{MH} \cos \theta$
4 $\mathrm{MH}(1-\cos \theta)$
Explanation:
D We know that, work dW is done against the torque in rotating a magnet by $\mathrm{d} \theta$ in a magnetic field. $\mathrm{dW}=\tau \mathrm{d} \theta=\mathrm{MH} \sin \theta \times \mathrm{d} \theta$ Where, $\mathrm{M}=$ Magnetic moment of magnet $\mathrm{H}=$ External magnetic field $\therefore$ Total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is given as $\mathrm{W}=\mathrm{MH}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ Since, Initially a dipole is aligned with magnetic field $\theta_{1}$ $=0^{\circ}$ and after rotating through angle $\theta_{2}=\theta$ Hence, total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is $\mathrm{W}=\mathrm{MH}(1-\cos \theta)$
UPSEE - 2005
Moving Charges & Magnetism
153875
Two identical magnetic dipoles of magnetic moment $2 \mathrm{Am}^{2}$ are placed at a separation of 2 $m$ with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is
1 $4 \sqrt{5} \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \sqrt{5} \times 10^{-7} \mathrm{~T}$
4 $2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
5 $4 \sqrt{2} \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, magnetic moment $(\mathrm{m})=2 \mathrm{Am}^{2}$ Since, the point $\mathrm{P}$ lies on the axial line of magnet $\mathrm{N}_{1} \mathrm{~S}_{1}$ at distance $\mathrm{r}_{1}=1 \mathrm{~m}$ and an equatorial line of magnet $\mathrm{N}_{2} \mathrm{~S}_{2}$ at distance $\mathrm{r}_{2}=1 \mathrm{~m}$ $\therefore \mathrm{B}_{1} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}_{1}}{\mathrm{r}_{1}^{3}}=10^{-7} \times \frac{2 \times 2}{1^{3}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{2} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{M}_{2}}{\mathrm{r}_{2}^{3}}=10^{-7} \times \frac{2}{1^{3}}=2 \times 10^{-7} \mathrm{~T}$ Hence, the resultant magnetic field at $\mathrm{P}$ $\mathrm{B}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}}$ $\mathrm{~B}=\sqrt{\left(4 \times 10^{-7}\right)^{2}+\left(2 \times 10^{-7}\right)^{2}}$ $\mathrm{~B}=2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
Kerala CEE - 2011
Moving Charges & Magnetism
153876
The oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is $0.5 \mathrm{~m}$, the kinetic energy of a proton, which is accelerated by the cyclotron is
1 $10.2 \mathrm{MeV}$
2 $2.55 \mathrm{MeV}$
3 $20.4 \mathrm{MeV}$
4 $5.1 \mathrm{MeV}$
5 $1.5 \mathrm{MeV}$
Explanation:
D Given, $\mathrm{f}=10 \mathrm{MHz}=10 \times 10^{6} \mathrm{~Hz}, \mathrm{r}=0.5 \mathrm{~m}$ We know that, $K . E=\frac{q^{2} B^{2} r^{2}}{2 m}$ And frequency $(f)=\frac{q B}{2 \pi m}$ or $\quad \mathrm{qB}=2 \pi \mathrm{mf}$ Putting these value in equation (i), we get- $\mathrm{K} . \mathrm{E}= \frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}=\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{f}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}$ $=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$ $=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$ $=8.23 \times 10^{-13} \text { Joule }$ $\mathrm{K} . \mathrm{E}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}$ $\mathrm{K} . \mathrm{E}=5.1 \mathrm{MeV}$
Kerala CEE - 2008
Moving Charges & Magnetism
153880
Assertion: Cyclotron is a device which is used to accelerate the positive ion. Reason: Cyclotron frequency depends upon the velocity.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
2 (c.) If the Assertion is correct but Reason is incorrect.
3 If both the Assertion and Reason are incorrect.
4 If the Assertion is incorrect but the Reason is correct.
Explanation:
C Cyclotron is utilized to accelerate the positive ion and cyclotron frequency $(f)=\frac{B q}{2 \pi m}$ Hence, cyclotron frequency does not depend upon velocity.
153873
If a magnetic dipole of dipole moment $M$ rotated through an angle $\theta$ with respect to the direction of the field $H$, then the work done is :
1 $\mathrm{MH} \sin \theta$
2 $\mathrm{MH}(1-\sin \theta)$
3 $\mathrm{MH} \cos \theta$
4 $\mathrm{MH}(1-\cos \theta)$
Explanation:
D We know that, work dW is done against the torque in rotating a magnet by $\mathrm{d} \theta$ in a magnetic field. $\mathrm{dW}=\tau \mathrm{d} \theta=\mathrm{MH} \sin \theta \times \mathrm{d} \theta$ Where, $\mathrm{M}=$ Magnetic moment of magnet $\mathrm{H}=$ External magnetic field $\therefore$ Total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is given as $\mathrm{W}=\mathrm{MH}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ Since, Initially a dipole is aligned with magnetic field $\theta_{1}$ $=0^{\circ}$ and after rotating through angle $\theta_{2}=\theta$ Hence, total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is $\mathrm{W}=\mathrm{MH}(1-\cos \theta)$
UPSEE - 2005
Moving Charges & Magnetism
153875
Two identical magnetic dipoles of magnetic moment $2 \mathrm{Am}^{2}$ are placed at a separation of 2 $m$ with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is
1 $4 \sqrt{5} \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \sqrt{5} \times 10^{-7} \mathrm{~T}$
4 $2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
5 $4 \sqrt{2} \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, magnetic moment $(\mathrm{m})=2 \mathrm{Am}^{2}$ Since, the point $\mathrm{P}$ lies on the axial line of magnet $\mathrm{N}_{1} \mathrm{~S}_{1}$ at distance $\mathrm{r}_{1}=1 \mathrm{~m}$ and an equatorial line of magnet $\mathrm{N}_{2} \mathrm{~S}_{2}$ at distance $\mathrm{r}_{2}=1 \mathrm{~m}$ $\therefore \mathrm{B}_{1} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}_{1}}{\mathrm{r}_{1}^{3}}=10^{-7} \times \frac{2 \times 2}{1^{3}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{2} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{M}_{2}}{\mathrm{r}_{2}^{3}}=10^{-7} \times \frac{2}{1^{3}}=2 \times 10^{-7} \mathrm{~T}$ Hence, the resultant magnetic field at $\mathrm{P}$ $\mathrm{B}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}}$ $\mathrm{~B}=\sqrt{\left(4 \times 10^{-7}\right)^{2}+\left(2 \times 10^{-7}\right)^{2}}$ $\mathrm{~B}=2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
Kerala CEE - 2011
Moving Charges & Magnetism
153876
The oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is $0.5 \mathrm{~m}$, the kinetic energy of a proton, which is accelerated by the cyclotron is
1 $10.2 \mathrm{MeV}$
2 $2.55 \mathrm{MeV}$
3 $20.4 \mathrm{MeV}$
4 $5.1 \mathrm{MeV}$
5 $1.5 \mathrm{MeV}$
Explanation:
D Given, $\mathrm{f}=10 \mathrm{MHz}=10 \times 10^{6} \mathrm{~Hz}, \mathrm{r}=0.5 \mathrm{~m}$ We know that, $K . E=\frac{q^{2} B^{2} r^{2}}{2 m}$ And frequency $(f)=\frac{q B}{2 \pi m}$ or $\quad \mathrm{qB}=2 \pi \mathrm{mf}$ Putting these value in equation (i), we get- $\mathrm{K} . \mathrm{E}= \frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}=\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{f}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}$ $=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$ $=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$ $=8.23 \times 10^{-13} \text { Joule }$ $\mathrm{K} . \mathrm{E}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}$ $\mathrm{K} . \mathrm{E}=5.1 \mathrm{MeV}$
Kerala CEE - 2008
Moving Charges & Magnetism
153880
Assertion: Cyclotron is a device which is used to accelerate the positive ion. Reason: Cyclotron frequency depends upon the velocity.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
2 (c.) If the Assertion is correct but Reason is incorrect.
3 If both the Assertion and Reason are incorrect.
4 If the Assertion is incorrect but the Reason is correct.
Explanation:
C Cyclotron is utilized to accelerate the positive ion and cyclotron frequency $(f)=\frac{B q}{2 \pi m}$ Hence, cyclotron frequency does not depend upon velocity.
153873
If a magnetic dipole of dipole moment $M$ rotated through an angle $\theta$ with respect to the direction of the field $H$, then the work done is :
1 $\mathrm{MH} \sin \theta$
2 $\mathrm{MH}(1-\sin \theta)$
3 $\mathrm{MH} \cos \theta$
4 $\mathrm{MH}(1-\cos \theta)$
Explanation:
D We know that, work dW is done against the torque in rotating a magnet by $\mathrm{d} \theta$ in a magnetic field. $\mathrm{dW}=\tau \mathrm{d} \theta=\mathrm{MH} \sin \theta \times \mathrm{d} \theta$ Where, $\mathrm{M}=$ Magnetic moment of magnet $\mathrm{H}=$ External magnetic field $\therefore$ Total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is given as $\mathrm{W}=\mathrm{MH}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ Since, Initially a dipole is aligned with magnetic field $\theta_{1}$ $=0^{\circ}$ and after rotating through angle $\theta_{2}=\theta$ Hence, total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is $\mathrm{W}=\mathrm{MH}(1-\cos \theta)$
UPSEE - 2005
Moving Charges & Magnetism
153875
Two identical magnetic dipoles of magnetic moment $2 \mathrm{Am}^{2}$ are placed at a separation of 2 $m$ with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is
1 $4 \sqrt{5} \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \sqrt{5} \times 10^{-7} \mathrm{~T}$
4 $2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
5 $4 \sqrt{2} \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, magnetic moment $(\mathrm{m})=2 \mathrm{Am}^{2}$ Since, the point $\mathrm{P}$ lies on the axial line of magnet $\mathrm{N}_{1} \mathrm{~S}_{1}$ at distance $\mathrm{r}_{1}=1 \mathrm{~m}$ and an equatorial line of magnet $\mathrm{N}_{2} \mathrm{~S}_{2}$ at distance $\mathrm{r}_{2}=1 \mathrm{~m}$ $\therefore \mathrm{B}_{1} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}_{1}}{\mathrm{r}_{1}^{3}}=10^{-7} \times \frac{2 \times 2}{1^{3}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{2} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{M}_{2}}{\mathrm{r}_{2}^{3}}=10^{-7} \times \frac{2}{1^{3}}=2 \times 10^{-7} \mathrm{~T}$ Hence, the resultant magnetic field at $\mathrm{P}$ $\mathrm{B}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}}$ $\mathrm{~B}=\sqrt{\left(4 \times 10^{-7}\right)^{2}+\left(2 \times 10^{-7}\right)^{2}}$ $\mathrm{~B}=2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
Kerala CEE - 2011
Moving Charges & Magnetism
153876
The oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is $0.5 \mathrm{~m}$, the kinetic energy of a proton, which is accelerated by the cyclotron is
1 $10.2 \mathrm{MeV}$
2 $2.55 \mathrm{MeV}$
3 $20.4 \mathrm{MeV}$
4 $5.1 \mathrm{MeV}$
5 $1.5 \mathrm{MeV}$
Explanation:
D Given, $\mathrm{f}=10 \mathrm{MHz}=10 \times 10^{6} \mathrm{~Hz}, \mathrm{r}=0.5 \mathrm{~m}$ We know that, $K . E=\frac{q^{2} B^{2} r^{2}}{2 m}$ And frequency $(f)=\frac{q B}{2 \pi m}$ or $\quad \mathrm{qB}=2 \pi \mathrm{mf}$ Putting these value in equation (i), we get- $\mathrm{K} . \mathrm{E}= \frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}=\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{f}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}$ $=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$ $=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$ $=8.23 \times 10^{-13} \text { Joule }$ $\mathrm{K} . \mathrm{E}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}$ $\mathrm{K} . \mathrm{E}=5.1 \mathrm{MeV}$
Kerala CEE - 2008
Moving Charges & Magnetism
153880
Assertion: Cyclotron is a device which is used to accelerate the positive ion. Reason: Cyclotron frequency depends upon the velocity.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
2 (c.) If the Assertion is correct but Reason is incorrect.
3 If both the Assertion and Reason are incorrect.
4 If the Assertion is incorrect but the Reason is correct.
Explanation:
C Cyclotron is utilized to accelerate the positive ion and cyclotron frequency $(f)=\frac{B q}{2 \pi m}$ Hence, cyclotron frequency does not depend upon velocity.
153873
If a magnetic dipole of dipole moment $M$ rotated through an angle $\theta$ with respect to the direction of the field $H$, then the work done is :
1 $\mathrm{MH} \sin \theta$
2 $\mathrm{MH}(1-\sin \theta)$
3 $\mathrm{MH} \cos \theta$
4 $\mathrm{MH}(1-\cos \theta)$
Explanation:
D We know that, work dW is done against the torque in rotating a magnet by $\mathrm{d} \theta$ in a magnetic field. $\mathrm{dW}=\tau \mathrm{d} \theta=\mathrm{MH} \sin \theta \times \mathrm{d} \theta$ Where, $\mathrm{M}=$ Magnetic moment of magnet $\mathrm{H}=$ External magnetic field $\therefore$ Total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is given as $\mathrm{W}=\mathrm{MH}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ Since, Initially a dipole is aligned with magnetic field $\theta_{1}$ $=0^{\circ}$ and after rotating through angle $\theta_{2}=\theta$ Hence, total work done in rotating the magnet from $\theta_{1}$ to $\theta_{2}$ is $\mathrm{W}=\mathrm{MH}(1-\cos \theta)$
UPSEE - 2005
Moving Charges & Magnetism
153875
Two identical magnetic dipoles of magnetic moment $2 \mathrm{Am}^{2}$ are placed at a separation of 2 $m$ with their axes perpendicular to each other in air. The resultant magnetic field at a midpoint between the dipoles is
1 $4 \sqrt{5} \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \sqrt{5} \times 10^{-7} \mathrm{~T}$
4 $2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
5 $4 \sqrt{2} \times 10^{-7} \mathrm{~T}$
Explanation:
D Given, magnetic moment $(\mathrm{m})=2 \mathrm{Am}^{2}$ Since, the point $\mathrm{P}$ lies on the axial line of magnet $\mathrm{N}_{1} \mathrm{~S}_{1}$ at distance $\mathrm{r}_{1}=1 \mathrm{~m}$ and an equatorial line of magnet $\mathrm{N}_{2} \mathrm{~S}_{2}$ at distance $\mathrm{r}_{2}=1 \mathrm{~m}$ $\therefore \mathrm{B}_{1} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{2 \mathrm{M}_{1}}{\mathrm{r}_{1}^{3}}=10^{-7} \times \frac{2 \times 2}{1^{3}}=4 \times 10^{-7} \mathrm{~T}$ $\mathrm{~B}_{2} =\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{M}_{2}}{\mathrm{r}_{2}^{3}}=10^{-7} \times \frac{2}{1^{3}}=2 \times 10^{-7} \mathrm{~T}$ Hence, the resultant magnetic field at $\mathrm{P}$ $\mathrm{B}=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}}$ $\mathrm{~B}=\sqrt{\left(4 \times 10^{-7}\right)^{2}+\left(2 \times 10^{-7}\right)^{2}}$ $\mathrm{~B}=2 \sqrt{5} \times 10^{-7} \mathrm{~T}$
Kerala CEE - 2011
Moving Charges & Magnetism
153876
The oscillating frequency of a cyclotron is 10 MHz. If the radius of its dees is $0.5 \mathrm{~m}$, the kinetic energy of a proton, which is accelerated by the cyclotron is
1 $10.2 \mathrm{MeV}$
2 $2.55 \mathrm{MeV}$
3 $20.4 \mathrm{MeV}$
4 $5.1 \mathrm{MeV}$
5 $1.5 \mathrm{MeV}$
Explanation:
D Given, $\mathrm{f}=10 \mathrm{MHz}=10 \times 10^{6} \mathrm{~Hz}, \mathrm{r}=0.5 \mathrm{~m}$ We know that, $K . E=\frac{q^{2} B^{2} r^{2}}{2 m}$ And frequency $(f)=\frac{q B}{2 \pi m}$ or $\quad \mathrm{qB}=2 \pi \mathrm{mf}$ Putting these value in equation (i), we get- $\mathrm{K} . \mathrm{E}= \frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}=\frac{4 \pi^{2} \mathrm{~m}^{2} \mathrm{f}^{2} \mathrm{r}^{2}}{2 \mathrm{~m}}$ $=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$ $=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$ $=8.23 \times 10^{-13} \text { Joule }$ $\mathrm{K} . \mathrm{E}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}$ $\mathrm{K} . \mathrm{E}=5.1 \mathrm{MeV}$
Kerala CEE - 2008
Moving Charges & Magnetism
153880
Assertion: Cyclotron is a device which is used to accelerate the positive ion. Reason: Cyclotron frequency depends upon the velocity.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
2 (c.) If the Assertion is correct but Reason is incorrect.
3 If both the Assertion and Reason are incorrect.
4 If the Assertion is incorrect but the Reason is correct.
Explanation:
C Cyclotron is utilized to accelerate the positive ion and cyclotron frequency $(f)=\frac{B q}{2 \pi m}$ Hence, cyclotron frequency does not depend upon velocity.
