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Serjik [45]
3 years ago
10

Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi

th a magnetic field of 0.400 t.
Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

The wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.400 t is 49.19

<h3>Explanation: </h3>

Electromagnetic radiation is the electromagnetic field waves that propagates through space and carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, light, ultraviolet, X-rays, and gamma rays.

A proton is a subatomic particle with symbol p or p⁺ with positive electric charge of +1e elementary charge and a mass slightly less than neutron.  

A cyclotron  a type of particle accelerator that accelerates charged particles outwards from the center along a spiral path.

Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.400 t.

  • B = magnetic field in the cyclotron = 0.400 T
  • q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C
  • m = mass of the proton = 1.67 x 10⁻²⁷ kg
  • f = frequency of revolution of proton in the cyclotron
  • v = speed of electromagnetic waves = 3 x 10⁸ m/s

\omega=\frac{v}{\rho} = \frac{qB}{m} =  \frac{1.6*10^{-19}C * 0.4t}{1.67*10^{-27} kg}\\\omega = \frac{0.64*10^{-19} Ct }{1.67*10^{-27} kg} = 0.383 * 10^{8}

T = \frac{2*\pi}{\omega} = \frac{2*\pi}{0.383*10^{8}} = 16.396*10^{-8}

\omega = \frac{2\pi c}{\lambda} \\ 0.383*10^8=  \frac{2\pi * 3*10^{8}}{\lambda} \\\lambda = \frac{18.84}{0.383} = 49.19

Learn more about   electromagnetic waves brainly.com/question/1750216

#LearnWithBrainly

levacccp [35]3 years ago
7 0

B = magnetic field in the cyclotron = 0.400 T

q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

f = frequency of revolution of proton in the cyclotron = ?

v = speed of electromagnetic waves = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = ?

Frequency of revolution of proton in the cyclotron is given as

f = qB/(2πm)

inserting the values

f = (1.6 x 10⁻¹⁹)(0.400)/(2 (3.14) (1.67 x 10⁻²⁷))

f = 6.1 x 10⁶ Hz

wavelength of electromagnetic wave is given as

λ = v/f

λ = (3 x 10⁸)/(6.1 x 10⁶)

λ = 49.2 m

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2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗
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Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  T =  48.255 \ N

b

   The time taken is  t =  0.226 \ s

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  V_{max} =0.384 \ m/s

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is m_b  =  5 \ kg

     The angle is  \theta = 10^o

     The length of the string is  L =  0.5 \ m

The tension on the string is mathematically represented as

              T  = mg cos \theta

substituting values

             T =  5 * 9.8 cos(10)

             T =  48.255 \ N

The motion of the bob is mathematically represented as

             S =  A sin (w t  + \frac{\pi }{2} )

     =>   S =  A sin (wt)

Where  w is the angular speed

and  \frac{\pi}{2} is the phase change

At initial position S =  0

   So   wt  = cos^{-1} (0)

          wt  = 1

Generally w can be mathematically represented as

          w =  \frac{2 \pi }{T}

Where T is the period of oscillation which i mathematically represented as

          T  =  2 \pi \sqrt{\frac{L}{g} }

So  

      t   = \frac{1}{w}

       t   = \frac{T}{2 \pi}

       t =  \sqrt{\frac{L}{g} }

substituting values

       t =  \sqrt{\frac{0.5}{9.8} }

       t =  0.226 \ s

Looking at the equation

         wt =  1

We see that maximum velocity of the bob will be at  S = 0  

i. e     w =  \frac{1}{t}

The maximum velocity is mathematically represented as

          V_{max}   =  w A

Where A is the amplitude which is mathematically represented as

         A = L sin \theta

So

      V_{max} = \frac{2 \pi }{T }  L sin \theta

      V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} }   L sin \theta  Recall   T  =  2 \pi \sqrt{\frac{L}{g} }

     V_{max} = \sqrt{gL} sin \theta

substituting values        

       V_{max} = \sqrt{9.8 * 0.5 } sin (10)

       V_{max} =0.384 \ m/s

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