NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153868
A cyclotron's oscillator frequency is $10 \mathrm{MHz}$. What should be the operating magnetic field for accelerating protons? $\left(\right.$ mass of the proton $\left.=1.67 \times 10^{-27} \mathbf{k g}\right)$
1 $0.33 \mathrm{~T}$
2 $0.66 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $3.0 \mathrm{~T}$
Explanation:
B Given, $\mathrm{f}=10 \mathrm{MHz}=10^{7} \mathrm{~Hz}, \mathrm{~m}=1.67 \times 10^{-}$ ${ }^{27} \mathrm{~kg}$. Charge on proton $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic field $(B)=\frac{2 \pi \mathrm{mf}}{\mathrm{q}}$ $\mathrm{B}=\mathrm{B}=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^{7}}{1.6 \times 10^{-19}}$ $\mathrm{~B}=0.66 \mathrm{~T}$
AMU-2018
Moving Charges & Magnetism
153877
In the cyclotron, as radius of the circular path of the charged particle increases : ( $\omega=$ angular velocity, $v=$ linear velocity)
1 both $\omega$ and $v$ increase
2 $\omega$ only increases, $v$ remains constant
3 $v$ increases, $\omega$ remains constant
4 $\mathrm{v}$ increases, $\omega$ decreases
Explanation:
C We know that, a radius of circular path is given by, $\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ $\mathrm{v}=\frac{\mathrm{qBR}}{\mathrm{m}}$ And Angular velocity $(\omega)=\frac{\mathrm{qB}}{\mathrm{m}}$ Therefore, $\mathrm{R} \propto \mathrm{v}$ As the radius increases, the velocity (v) of the particles will also increase. And the angular velocity $(\omega)$ is independent of the radius of the circular path therefore, it will remain same or constant.
Karnataka CET-2016
Moving Charges & Magnetism
153879
Identify the correct statement from the following.
1 Cyclotron frequency is dependent on speed of the charged particle.
2 Kinetic energy of charged particle in cyclotron does not depend on its mass.
3 Cyclotron frequency does not depend on speed of the charged particle.
4 Kinetic energy of charged particle in cyclotron is independent of its charge.
Explanation:
C Cyclotron frequency depends upon the charge present on the particle. The frequency of the cyclotron depends upon the magnetic field that influences the acceleration of the charged particle. Since, frequency $(\mathrm{f})=\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}$ Hence, from the above relation we can conclude that the cyclotron frequency does not depend upon the speed of the charged particle.
J and K CET- 2008
Moving Charges & Magnetism
153883
In cyclotron, for a given magnet, radius of the semicircle traced by positive ion is directly proportional to $(v=$ velocity of positive ion)
1 $\mathrm{v}^{-2}$
2 $\mathrm{v}^{-1}$
3 $\mathrm{V}$
4 $\mathrm{v}^{2}$
Explanation:
C We know that, magnetic force only provides the centripetal force- $\mathrm{Bqv} =\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{R} =\frac{\mathrm{mv}}{\mathrm{Bq}}$ $\mathrm{R} \propto \mathrm{v}$
153868
A cyclotron's oscillator frequency is $10 \mathrm{MHz}$. What should be the operating magnetic field for accelerating protons? $\left(\right.$ mass of the proton $\left.=1.67 \times 10^{-27} \mathbf{k g}\right)$
1 $0.33 \mathrm{~T}$
2 $0.66 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $3.0 \mathrm{~T}$
Explanation:
B Given, $\mathrm{f}=10 \mathrm{MHz}=10^{7} \mathrm{~Hz}, \mathrm{~m}=1.67 \times 10^{-}$ ${ }^{27} \mathrm{~kg}$. Charge on proton $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic field $(B)=\frac{2 \pi \mathrm{mf}}{\mathrm{q}}$ $\mathrm{B}=\mathrm{B}=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^{7}}{1.6 \times 10^{-19}}$ $\mathrm{~B}=0.66 \mathrm{~T}$
AMU-2018
Moving Charges & Magnetism
153877
In the cyclotron, as radius of the circular path of the charged particle increases : ( $\omega=$ angular velocity, $v=$ linear velocity)
1 both $\omega$ and $v$ increase
2 $\omega$ only increases, $v$ remains constant
3 $v$ increases, $\omega$ remains constant
4 $\mathrm{v}$ increases, $\omega$ decreases
Explanation:
C We know that, a radius of circular path is given by, $\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ $\mathrm{v}=\frac{\mathrm{qBR}}{\mathrm{m}}$ And Angular velocity $(\omega)=\frac{\mathrm{qB}}{\mathrm{m}}$ Therefore, $\mathrm{R} \propto \mathrm{v}$ As the radius increases, the velocity (v) of the particles will also increase. And the angular velocity $(\omega)$ is independent of the radius of the circular path therefore, it will remain same or constant.
Karnataka CET-2016
Moving Charges & Magnetism
153879
Identify the correct statement from the following.
1 Cyclotron frequency is dependent on speed of the charged particle.
2 Kinetic energy of charged particle in cyclotron does not depend on its mass.
3 Cyclotron frequency does not depend on speed of the charged particle.
4 Kinetic energy of charged particle in cyclotron is independent of its charge.
Explanation:
C Cyclotron frequency depends upon the charge present on the particle. The frequency of the cyclotron depends upon the magnetic field that influences the acceleration of the charged particle. Since, frequency $(\mathrm{f})=\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}$ Hence, from the above relation we can conclude that the cyclotron frequency does not depend upon the speed of the charged particle.
J and K CET- 2008
Moving Charges & Magnetism
153883
In cyclotron, for a given magnet, radius of the semicircle traced by positive ion is directly proportional to $(v=$ velocity of positive ion)
1 $\mathrm{v}^{-2}$
2 $\mathrm{v}^{-1}$
3 $\mathrm{V}$
4 $\mathrm{v}^{2}$
Explanation:
C We know that, magnetic force only provides the centripetal force- $\mathrm{Bqv} =\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{R} =\frac{\mathrm{mv}}{\mathrm{Bq}}$ $\mathrm{R} \propto \mathrm{v}$
153868
A cyclotron's oscillator frequency is $10 \mathrm{MHz}$. What should be the operating magnetic field for accelerating protons? $\left(\right.$ mass of the proton $\left.=1.67 \times 10^{-27} \mathbf{k g}\right)$
1 $0.33 \mathrm{~T}$
2 $0.66 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $3.0 \mathrm{~T}$
Explanation:
B Given, $\mathrm{f}=10 \mathrm{MHz}=10^{7} \mathrm{~Hz}, \mathrm{~m}=1.67 \times 10^{-}$ ${ }^{27} \mathrm{~kg}$. Charge on proton $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic field $(B)=\frac{2 \pi \mathrm{mf}}{\mathrm{q}}$ $\mathrm{B}=\mathrm{B}=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^{7}}{1.6 \times 10^{-19}}$ $\mathrm{~B}=0.66 \mathrm{~T}$
AMU-2018
Moving Charges & Magnetism
153877
In the cyclotron, as radius of the circular path of the charged particle increases : ( $\omega=$ angular velocity, $v=$ linear velocity)
1 both $\omega$ and $v$ increase
2 $\omega$ only increases, $v$ remains constant
3 $v$ increases, $\omega$ remains constant
4 $\mathrm{v}$ increases, $\omega$ decreases
Explanation:
C We know that, a radius of circular path is given by, $\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ $\mathrm{v}=\frac{\mathrm{qBR}}{\mathrm{m}}$ And Angular velocity $(\omega)=\frac{\mathrm{qB}}{\mathrm{m}}$ Therefore, $\mathrm{R} \propto \mathrm{v}$ As the radius increases, the velocity (v) of the particles will also increase. And the angular velocity $(\omega)$ is independent of the radius of the circular path therefore, it will remain same or constant.
Karnataka CET-2016
Moving Charges & Magnetism
153879
Identify the correct statement from the following.
1 Cyclotron frequency is dependent on speed of the charged particle.
2 Kinetic energy of charged particle in cyclotron does not depend on its mass.
3 Cyclotron frequency does not depend on speed of the charged particle.
4 Kinetic energy of charged particle in cyclotron is independent of its charge.
Explanation:
C Cyclotron frequency depends upon the charge present on the particle. The frequency of the cyclotron depends upon the magnetic field that influences the acceleration of the charged particle. Since, frequency $(\mathrm{f})=\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}$ Hence, from the above relation we can conclude that the cyclotron frequency does not depend upon the speed of the charged particle.
J and K CET- 2008
Moving Charges & Magnetism
153883
In cyclotron, for a given magnet, radius of the semicircle traced by positive ion is directly proportional to $(v=$ velocity of positive ion)
1 $\mathrm{v}^{-2}$
2 $\mathrm{v}^{-1}$
3 $\mathrm{V}$
4 $\mathrm{v}^{2}$
Explanation:
C We know that, magnetic force only provides the centripetal force- $\mathrm{Bqv} =\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{R} =\frac{\mathrm{mv}}{\mathrm{Bq}}$ $\mathrm{R} \propto \mathrm{v}$
153868
A cyclotron's oscillator frequency is $10 \mathrm{MHz}$. What should be the operating magnetic field for accelerating protons? $\left(\right.$ mass of the proton $\left.=1.67 \times 10^{-27} \mathbf{k g}\right)$
1 $0.33 \mathrm{~T}$
2 $0.66 \mathrm{~T}$
3 $1.5 \mathrm{~T}$
4 $3.0 \mathrm{~T}$
Explanation:
B Given, $\mathrm{f}=10 \mathrm{MHz}=10^{7} \mathrm{~Hz}, \mathrm{~m}=1.67 \times 10^{-}$ ${ }^{27} \mathrm{~kg}$. Charge on proton $(\mathrm{q})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic field $(B)=\frac{2 \pi \mathrm{mf}}{\mathrm{q}}$ $\mathrm{B}=\mathrm{B}=\frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^{7}}{1.6 \times 10^{-19}}$ $\mathrm{~B}=0.66 \mathrm{~T}$
AMU-2018
Moving Charges & Magnetism
153877
In the cyclotron, as radius of the circular path of the charged particle increases : ( $\omega=$ angular velocity, $v=$ linear velocity)
1 both $\omega$ and $v$ increase
2 $\omega$ only increases, $v$ remains constant
3 $v$ increases, $\omega$ remains constant
4 $\mathrm{v}$ increases, $\omega$ decreases
Explanation:
C We know that, a radius of circular path is given by, $\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ $\mathrm{v}=\frac{\mathrm{qBR}}{\mathrm{m}}$ And Angular velocity $(\omega)=\frac{\mathrm{qB}}{\mathrm{m}}$ Therefore, $\mathrm{R} \propto \mathrm{v}$ As the radius increases, the velocity (v) of the particles will also increase. And the angular velocity $(\omega)$ is independent of the radius of the circular path therefore, it will remain same or constant.
Karnataka CET-2016
Moving Charges & Magnetism
153879
Identify the correct statement from the following.
1 Cyclotron frequency is dependent on speed of the charged particle.
2 Kinetic energy of charged particle in cyclotron does not depend on its mass.
3 Cyclotron frequency does not depend on speed of the charged particle.
4 Kinetic energy of charged particle in cyclotron is independent of its charge.
Explanation:
C Cyclotron frequency depends upon the charge present on the particle. The frequency of the cyclotron depends upon the magnetic field that influences the acceleration of the charged particle. Since, frequency $(\mathrm{f})=\frac{\mathrm{Bq}}{2 \pi \mathrm{m}}$ Hence, from the above relation we can conclude that the cyclotron frequency does not depend upon the speed of the charged particle.
J and K CET- 2008
Moving Charges & Magnetism
153883
In cyclotron, for a given magnet, radius of the semicircle traced by positive ion is directly proportional to $(v=$ velocity of positive ion)
1 $\mathrm{v}^{-2}$
2 $\mathrm{v}^{-1}$
3 $\mathrm{V}$
4 $\mathrm{v}^{2}$
Explanation:
C We know that, magnetic force only provides the centripetal force- $\mathrm{Bqv} =\frac{\mathrm{mv}^{2}}{\mathrm{R}}$ $\mathrm{R} =\frac{\mathrm{mv}}{\mathrm{Bq}}$ $\mathrm{R} \propto \mathrm{v}$
