The oscillation frequency of a cyclotron is $10 MHz$ . What should be the operating magnetic field for accelerating proton?

0.156 T

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0.256 T

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0.356 T

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0.656 T

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Solution

The correct option is D $0.656 T$
Frequency, $f = \frac{q B}{2 π m}$

Mass of proton, $m = 1.67 \times 10^{- 27} kg$
Charge on proton, $q = 1.6 \times 10^{- 19} C$

$∴ 10 \times 10^{6} = \frac{1.6 \times 10^{- 19} \times B}{2 π \times 1.67 \times 10^{- 27}}$

$\Rightarrow B \approx 0.656 T$

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Similar questions

Q. The oscillator frequency of a cyclotron is 10 MHz what should be the operating magnetic field accelerating proton?

Q. The approximate operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of

12 MHz

is:

A cyclotron’s oscillator frequency is 10 MHz. the operating magnetic field for the cyclotron to accelerate the proton is ( mass of proton $= 1.67 \times 10^{- 27} k g$

Q. The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is ( q =

1.6 \times 10^{- 19}

m_{p} = 1.67 \times 10^{- 27}

kg and 1 MeV =

1.6 \times 10^{- 13}

Q. The magnitude of magnetic field required to accelerate protons (mass =

1.67 \times 10^{- 27} k g

) in a cyclotron that is operated at an oscillator frequency 10 MHz is approximately

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The oscillation frequency of a cyclotron is 10 MHz. What should be the operating magnetic field for accelerating proton?

The correct option is D 0.656 TFrequency, f=qB2πm Mass of proton, m=1.67×10−27 kg Charge on proton, q=1.6×10−19 C ∴10×106=1.6×10−19×B2π×1.67×10−27 ⇒B≈0.656 T

The oscillation frequency of a cyclotron is $10 MHz$ . What should be the operating magnetic field for accelerating proton?

The correct option is D $0.656 T$
Frequency, $f = \frac{q B}{2 π m}$

Mass of proton, $m = 1.67 \times 10^{- 27} kg$
Charge on proton, $q = 1.6 \times 10^{- 19} C$

$∴ 10 \times 10^{6} = \frac{1.6 \times 10^{- 19} \times B}{2 π \times 1.67 \times 10^{- 27}}$

$\Rightarrow B \approx 0.656 T$