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PHXII04:MOVING CHARGES AND MAGNETISM

362783 A cyclotron's oscillator frequency is 10 $M H z$ and the operating magnetic fiel is 0.66 $T$ . If the radius of its dees is 60 $c m$ , then the kinetic energy of the proton beam produced by the accelerator is

1 9

M e V

2 10

M e V

3 7

M e V

4 11

M e V

Explanation:

Here,
Oscillator frequency, $v = 10 M H z = 10^{6} H z$
Radius of the dees, $R = 60 c m = 0.6 m$
Magnetic field, $B = 0.66 T$
Charge on the proton, $q = 1.6 \times 10^{- 19} C$
Mass of the proton, $m = 1.67 \times 10^{- 27} k g$
The kinetuc energy of the proton beam produced by the accelertor is
$K = \frac{B^{2} q^{2} R^{2}}{2 m}$
$= \frac{{(0.66 T)}^{2} {(1.6 \times 10^{- 19} C)}^{2} {(0.6 m)}^{2}}{2 (1.67 \times 10^{- 27} k g)}$
$= 1.2 \times 10^{- 12} J$
$= \frac{1.2 \times 10^{- 12}}{1.6 \times 10^{- 13}} M e V (as 1 M v = 1.6 \times 10^{- 13} J)$
$= 7.5 M e V = 7 M e V$

KCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362784 A cyclotron is used to accelerate protons $(_{1}^{1} H)$ , Deuterons $(_{1}^{2} H)$ and $α$ -particles $(_{2}^{4} H e)$ . While exiting under similar conditions, the minimum $K . E .$ is gained by

1 Deuteron

2 Same for all

3

α

-particle

4 Proton

Explanation:

For a cyclotron.
$K \cdot E_{max} = \frac{B^{2} q^{2} r_{0}^{2}}{2 m} \Rightarrow K \cdot E_{max} \propto \frac{q^{2}}{m}$
$∴$ Minimum K.E. is gained by deutron.

KCET - 2020

PHXII04:MOVING CHARGES AND MAGNETISM

362785 An alternating electric field of frequency $v$ , is applied across the dees (radius $= R$ ) of a cyclotron that is being used to accelerate protons (mass $= m$ ). The operating magnetic field ( $B$ ) used in the cyclotron and the kinetic energy ( $K$ ) of the proton beam, produced by it, are given by

1

B = \frac{2 π m v}{e}

and

K = m^{2} π v R^{2}

2

B = \frac{m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

3

B = \frac{m v}{e}

and

K = m^{2} π v R^{2}

4

B = \frac{2 π m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

Explanation:

Frequency, $f = \frac{e B}{2 π m}$
$K E = \frac{1}{2} m v^{2}$ and radius $R = \frac{m v}{e B}$
Here, velocity, $v = \frac{π R}{\frac{T}{2}} = \frac{2 π R}{T} = 2 π R f$
$∴$ Radius, $R = \frac{m (2 π R f)}{e B}$
Magnetic field, $B = \frac{2 π m f}{e}$
Kinetic energy, $K = \frac{1}{2} m (2 π R f)^{2}$
$= 2 m π^{2} f^{2} R^{2}$

MHTCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362786 Assertion :
Cyclotron does not accelerate electron.
Reason :
Mass of the electron is very small.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For same $F$ , we have acceleration $\propto (\frac{1}{mass})$ Cyclotrons are more effective for accelerating heavy particles like protons $(p)$ and alpha particles because the force depends on the charge-to-mass ratio. Electrons, have very small mass. So correct option is (1).

PHXII04:MOVING CHARGES AND MAGNETISM

362787 The oscillating frequency of a cyclotron is 10 $M H z$ . If the radius of dees is 0.5 $m$ , the maximum kinetic energy of a protons, which is accelerated by the cyclotron is

1

2.55 M e v

2

20.4 M e v

3

10.2 M e v

4

5.1 M e v

Explanation:

$K E = \frac{1}{2} m {(\frac{B q r_{0}}{m})}^{2} & v = \frac{B q}{2 π m}$
$K E = 5.1 M e V$

PHXII04:MOVING CHARGES AND MAGNETISM

362783 A cyclotron's oscillator frequency is 10 $M H z$ and the operating magnetic fiel is 0.66 $T$ . If the radius of its dees is 60 $c m$ , then the kinetic energy of the proton beam produced by the accelerator is

1 9

M e V

2 10

M e V

3 7

M e V

4 11

M e V

Explanation:

Here,
Oscillator frequency, $v = 10 M H z = 10^{6} H z$
Radius of the dees, $R = 60 c m = 0.6 m$
Magnetic field, $B = 0.66 T$
Charge on the proton, $q = 1.6 \times 10^{- 19} C$
Mass of the proton, $m = 1.67 \times 10^{- 27} k g$
The kinetuc energy of the proton beam produced by the accelertor is
$K = \frac{B^{2} q^{2} R^{2}}{2 m}$
$= \frac{{(0.66 T)}^{2} {(1.6 \times 10^{- 19} C)}^{2} {(0.6 m)}^{2}}{2 (1.67 \times 10^{- 27} k g)}$
$= 1.2 \times 10^{- 12} J$
$= \frac{1.2 \times 10^{- 12}}{1.6 \times 10^{- 13}} M e V (as 1 M v = 1.6 \times 10^{- 13} J)$
$= 7.5 M e V = 7 M e V$

KCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362784 A cyclotron is used to accelerate protons $(_{1}^{1} H)$ , Deuterons $(_{1}^{2} H)$ and $α$ -particles $(_{2}^{4} H e)$ . While exiting under similar conditions, the minimum $K . E .$ is gained by

1 Deuteron

2 Same for all

3

α

-particle

4 Proton

Explanation:

For a cyclotron.
$K \cdot E_{max} = \frac{B^{2} q^{2} r_{0}^{2}}{2 m} \Rightarrow K \cdot E_{max} \propto \frac{q^{2}}{m}$
$∴$ Minimum K.E. is gained by deutron.

KCET - 2020

PHXII04:MOVING CHARGES AND MAGNETISM

362785 An alternating electric field of frequency $v$ , is applied across the dees (radius $= R$ ) of a cyclotron that is being used to accelerate protons (mass $= m$ ). The operating magnetic field ( $B$ ) used in the cyclotron and the kinetic energy ( $K$ ) of the proton beam, produced by it, are given by

1

B = \frac{2 π m v}{e}

and

K = m^{2} π v R^{2}

2

B = \frac{m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

3

B = \frac{m v}{e}

and

K = m^{2} π v R^{2}

4

B = \frac{2 π m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

Explanation:

Frequency, $f = \frac{e B}{2 π m}$
$K E = \frac{1}{2} m v^{2}$ and radius $R = \frac{m v}{e B}$
Here, velocity, $v = \frac{π R}{\frac{T}{2}} = \frac{2 π R}{T} = 2 π R f$
$∴$ Radius, $R = \frac{m (2 π R f)}{e B}$
Magnetic field, $B = \frac{2 π m f}{e}$
Kinetic energy, $K = \frac{1}{2} m (2 π R f)^{2}$
$= 2 m π^{2} f^{2} R^{2}$

MHTCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362786 Assertion :
Cyclotron does not accelerate electron.
Reason :
Mass of the electron is very small.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For same $F$ , we have acceleration $\propto (\frac{1}{mass})$ Cyclotrons are more effective for accelerating heavy particles like protons $(p)$ and alpha particles because the force depends on the charge-to-mass ratio. Electrons, have very small mass. So correct option is (1).

PHXII04:MOVING CHARGES AND MAGNETISM

362787 The oscillating frequency of a cyclotron is 10 $M H z$ . If the radius of dees is 0.5 $m$ , the maximum kinetic energy of a protons, which is accelerated by the cyclotron is

1

2.55 M e v

2

20.4 M e v

3

10.2 M e v

4

5.1 M e v

Explanation:

$K E = \frac{1}{2} m {(\frac{B q r_{0}}{m})}^{2} & v = \frac{B q}{2 π m}$
$K E = 5.1 M e V$

PHXII04:MOVING CHARGES AND MAGNETISM

362783 A cyclotron's oscillator frequency is 10 $M H z$ and the operating magnetic fiel is 0.66 $T$ . If the radius of its dees is 60 $c m$ , then the kinetic energy of the proton beam produced by the accelerator is

1 9

M e V

2 10

M e V

3 7

M e V

4 11

M e V

Explanation:

Here,
Oscillator frequency, $v = 10 M H z = 10^{6} H z$
Radius of the dees, $R = 60 c m = 0.6 m$
Magnetic field, $B = 0.66 T$
Charge on the proton, $q = 1.6 \times 10^{- 19} C$
Mass of the proton, $m = 1.67 \times 10^{- 27} k g$
The kinetuc energy of the proton beam produced by the accelertor is
$K = \frac{B^{2} q^{2} R^{2}}{2 m}$
$= \frac{{(0.66 T)}^{2} {(1.6 \times 10^{- 19} C)}^{2} {(0.6 m)}^{2}}{2 (1.67 \times 10^{- 27} k g)}$
$= 1.2 \times 10^{- 12} J$
$= \frac{1.2 \times 10^{- 12}}{1.6 \times 10^{- 13}} M e V (as 1 M v = 1.6 \times 10^{- 13} J)$
$= 7.5 M e V = 7 M e V$

KCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362784 A cyclotron is used to accelerate protons $(_{1}^{1} H)$ , Deuterons $(_{1}^{2} H)$ and $α$ -particles $(_{2}^{4} H e)$ . While exiting under similar conditions, the minimum $K . E .$ is gained by

1 Deuteron

2 Same for all

3

α

-particle

4 Proton

Explanation:

For a cyclotron.
$K \cdot E_{max} = \frac{B^{2} q^{2} r_{0}^{2}}{2 m} \Rightarrow K \cdot E_{max} \propto \frac{q^{2}}{m}$
$∴$ Minimum K.E. is gained by deutron.

KCET - 2020

PHXII04:MOVING CHARGES AND MAGNETISM

362785 An alternating electric field of frequency $v$ , is applied across the dees (radius $= R$ ) of a cyclotron that is being used to accelerate protons (mass $= m$ ). The operating magnetic field ( $B$ ) used in the cyclotron and the kinetic energy ( $K$ ) of the proton beam, produced by it, are given by

1

B = \frac{2 π m v}{e}

and

K = m^{2} π v R^{2}

2

B = \frac{m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

3

B = \frac{m v}{e}

and

K = m^{2} π v R^{2}

4

B = \frac{2 π m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

Explanation:

Frequency, $f = \frac{e B}{2 π m}$
$K E = \frac{1}{2} m v^{2}$ and radius $R = \frac{m v}{e B}$
Here, velocity, $v = \frac{π R}{\frac{T}{2}} = \frac{2 π R}{T} = 2 π R f$
$∴$ Radius, $R = \frac{m (2 π R f)}{e B}$
Magnetic field, $B = \frac{2 π m f}{e}$
Kinetic energy, $K = \frac{1}{2} m (2 π R f)^{2}$
$= 2 m π^{2} f^{2} R^{2}$

MHTCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362786 Assertion :
Cyclotron does not accelerate electron.
Reason :
Mass of the electron is very small.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For same $F$ , we have acceleration $\propto (\frac{1}{mass})$ Cyclotrons are more effective for accelerating heavy particles like protons $(p)$ and alpha particles because the force depends on the charge-to-mass ratio. Electrons, have very small mass. So correct option is (1).

PHXII04:MOVING CHARGES AND MAGNETISM

362787 The oscillating frequency of a cyclotron is 10 $M H z$ . If the radius of dees is 0.5 $m$ , the maximum kinetic energy of a protons, which is accelerated by the cyclotron is

1

2.55 M e v

2

20.4 M e v

3

10.2 M e v

4

5.1 M e v

Explanation:

$K E = \frac{1}{2} m {(\frac{B q r_{0}}{m})}^{2} & v = \frac{B q}{2 π m}$
$K E = 5.1 M e V$

PHXII04:MOVING CHARGES AND MAGNETISM

362783 A cyclotron's oscillator frequency is 10 $M H z$ and the operating magnetic fiel is 0.66 $T$ . If the radius of its dees is 60 $c m$ , then the kinetic energy of the proton beam produced by the accelerator is

1 9

M e V

2 10

M e V

3 7

M e V

4 11

M e V

Explanation:

Here,
Oscillator frequency, $v = 10 M H z = 10^{6} H z$
Radius of the dees, $R = 60 c m = 0.6 m$
Magnetic field, $B = 0.66 T$
Charge on the proton, $q = 1.6 \times 10^{- 19} C$
Mass of the proton, $m = 1.67 \times 10^{- 27} k g$
The kinetuc energy of the proton beam produced by the accelertor is
$K = \frac{B^{2} q^{2} R^{2}}{2 m}$
$= \frac{{(0.66 T)}^{2} {(1.6 \times 10^{- 19} C)}^{2} {(0.6 m)}^{2}}{2 (1.67 \times 10^{- 27} k g)}$
$= 1.2 \times 10^{- 12} J$
$= \frac{1.2 \times 10^{- 12}}{1.6 \times 10^{- 13}} M e V (as 1 M v = 1.6 \times 10^{- 13} J)$
$= 7.5 M e V = 7 M e V$

KCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362784 A cyclotron is used to accelerate protons $(_{1}^{1} H)$ , Deuterons $(_{1}^{2} H)$ and $α$ -particles $(_{2}^{4} H e)$ . While exiting under similar conditions, the minimum $K . E .$ is gained by

1 Deuteron

2 Same for all

3

α

-particle

4 Proton

Explanation:

For a cyclotron.
$K \cdot E_{max} = \frac{B^{2} q^{2} r_{0}^{2}}{2 m} \Rightarrow K \cdot E_{max} \propto \frac{q^{2}}{m}$
$∴$ Minimum K.E. is gained by deutron.

KCET - 2020

PHXII04:MOVING CHARGES AND MAGNETISM

362785 An alternating electric field of frequency $v$ , is applied across the dees (radius $= R$ ) of a cyclotron that is being used to accelerate protons (mass $= m$ ). The operating magnetic field ( $B$ ) used in the cyclotron and the kinetic energy ( $K$ ) of the proton beam, produced by it, are given by

1

B = \frac{2 π m v}{e}

and

K = m^{2} π v R^{2}

2

B = \frac{m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

3

B = \frac{m v}{e}

and

K = m^{2} π v R^{2}

4

B = \frac{2 π m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

Explanation:

Frequency, $f = \frac{e B}{2 π m}$
$K E = \frac{1}{2} m v^{2}$ and radius $R = \frac{m v}{e B}$
Here, velocity, $v = \frac{π R}{\frac{T}{2}} = \frac{2 π R}{T} = 2 π R f$
$∴$ Radius, $R = \frac{m (2 π R f)}{e B}$
Magnetic field, $B = \frac{2 π m f}{e}$
Kinetic energy, $K = \frac{1}{2} m (2 π R f)^{2}$
$= 2 m π^{2} f^{2} R^{2}$

MHTCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362786 Assertion :
Cyclotron does not accelerate electron.
Reason :
Mass of the electron is very small.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For same $F$ , we have acceleration $\propto (\frac{1}{mass})$ Cyclotrons are more effective for accelerating heavy particles like protons $(p)$ and alpha particles because the force depends on the charge-to-mass ratio. Electrons, have very small mass. So correct option is (1).

PHXII04:MOVING CHARGES AND MAGNETISM

362787 The oscillating frequency of a cyclotron is 10 $M H z$ . If the radius of dees is 0.5 $m$ , the maximum kinetic energy of a protons, which is accelerated by the cyclotron is

1

2.55 M e v

2

20.4 M e v

3

10.2 M e v

4

5.1 M e v

Explanation:

$K E = \frac{1}{2} m {(\frac{B q r_{0}}{m})}^{2} & v = \frac{B q}{2 π m}$
$K E = 5.1 M e V$

PHXII04:MOVING CHARGES AND MAGNETISM

362783 A cyclotron's oscillator frequency is 10 $M H z$ and the operating magnetic fiel is 0.66 $T$ . If the radius of its dees is 60 $c m$ , then the kinetic energy of the proton beam produced by the accelerator is

1 9

M e V

2 10

M e V

3 7

M e V

4 11

M e V

Explanation:

Here,
Oscillator frequency, $v = 10 M H z = 10^{6} H z$
Radius of the dees, $R = 60 c m = 0.6 m$
Magnetic field, $B = 0.66 T$
Charge on the proton, $q = 1.6 \times 10^{- 19} C$
Mass of the proton, $m = 1.67 \times 10^{- 27} k g$
The kinetuc energy of the proton beam produced by the accelertor is
$K = \frac{B^{2} q^{2} R^{2}}{2 m}$
$= \frac{{(0.66 T)}^{2} {(1.6 \times 10^{- 19} C)}^{2} {(0.6 m)}^{2}}{2 (1.67 \times 10^{- 27} k g)}$
$= 1.2 \times 10^{- 12} J$
$= \frac{1.2 \times 10^{- 12}}{1.6 \times 10^{- 13}} M e V (as 1 M v = 1.6 \times 10^{- 13} J)$
$= 7.5 M e V = 7 M e V$

KCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362784 A cyclotron is used to accelerate protons $(_{1}^{1} H)$ , Deuterons $(_{1}^{2} H)$ and $α$ -particles $(_{2}^{4} H e)$ . While exiting under similar conditions, the minimum $K . E .$ is gained by

1 Deuteron

2 Same for all

3

α

-particle

4 Proton

Explanation:

For a cyclotron.
$K \cdot E_{max} = \frac{B^{2} q^{2} r_{0}^{2}}{2 m} \Rightarrow K \cdot E_{max} \propto \frac{q^{2}}{m}$
$∴$ Minimum K.E. is gained by deutron.

KCET - 2020

PHXII04:MOVING CHARGES AND MAGNETISM

362785 An alternating electric field of frequency $v$ , is applied across the dees (radius $= R$ ) of a cyclotron that is being used to accelerate protons (mass $= m$ ). The operating magnetic field ( $B$ ) used in the cyclotron and the kinetic energy ( $K$ ) of the proton beam, produced by it, are given by

1

B = \frac{2 π m v}{e}

and

K = m^{2} π v R^{2}

2

B = \frac{m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

3

B = \frac{m v}{e}

and

K = m^{2} π v R^{2}

4

B = \frac{2 π m v}{e}

and

K = 2 m π^{2} v^{2} R^{2}

Explanation:

Frequency, $f = \frac{e B}{2 π m}$
$K E = \frac{1}{2} m v^{2}$ and radius $R = \frac{m v}{e B}$
Here, velocity, $v = \frac{π R}{\frac{T}{2}} = \frac{2 π R}{T} = 2 π R f$
$∴$ Radius, $R = \frac{m (2 π R f)}{e B}$
Magnetic field, $B = \frac{2 π m f}{e}$
Kinetic energy, $K = \frac{1}{2} m (2 π R f)^{2}$
$= 2 m π^{2} f^{2} R^{2}$

MHTCET - 2018

PHXII04:MOVING CHARGES AND MAGNETISM

362786 Assertion :
Cyclotron does not accelerate electron.
Reason :
Mass of the electron is very small.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For same $F$ , we have acceleration $\propto (\frac{1}{mass})$ Cyclotrons are more effective for accelerating heavy particles like protons $(p)$ and alpha particles because the force depends on the charge-to-mass ratio. Electrons, have very small mass. So correct option is (1).

PHXII04:MOVING CHARGES AND MAGNETISM

362787 The oscillating frequency of a cyclotron is 10 $M H z$ . If the radius of dees is 0.5 $m$ , the maximum kinetic energy of a protons, which is accelerated by the cyclotron is

1

2.55 M e v

2

20.4 M e v

3

10.2 M e v

4

5.1 M e v

Explanation:

$K E = \frac{1}{2} m {(\frac{B q r_{0}}{m})}^{2} & v = \frac{B q}{2 π m}$
$K E = 5.1 M e V$

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