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Homework Help: Magnetic Fields and Proton Beam therapy

  1. Mar 22, 2010 #1
    Proton Beam therapy is an advanced technique to destroy tumors by concentrating a high energy beam of protons to a specific depth inside the body of a patient. These proton beams are produced by accelerators and steered to the patient-treatment rooms by magnets. In a particular cancer treatment center, the speed of the protons is 1.8 x 10^8 m/s. In a particular arrangement, the proton beam is deflected by 90 degrees with respect to its initial direction by a constant magnetic field.

    a) If the radius of the circular arc of the trajectory is 4.0 m, find the magnitude of the constant magnetic field.

    b) The engineers who installed these magnets tested the magnetic field by inserting a single square coil of side .2 m. The plane of the coil was made parallel to the field lines. They measured the torque on this coil when it carried a current. They found th torque to be .15 N.m for the magnetic field you found in part (a). What was the current in the square coil?

    For part A, I know I can use the equation:

    B = uI/2(pi)r

    Plug in what I know, I end up with :

    B = 4(pi)x10^-7 (I) / 2(Pi)(4.0m)


    2 (pi) x 10^-7 (I) / 4.0 m

    However I am unsure what the current is and therefore, do not know what to plug in for I.
  2. jcsd
  3. Mar 22, 2010 #2


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    Homework Helper

    B = uI/2(pi)r only applies to an infinitely long current-carrying wire. Here, you have to derive the equation yourself. What's the centripetal force equal to? What is it applied by? Also, since 1.8E8 m/s is more than half the speed of light, do you need to take into consideration relativistic effects?
  4. Mar 23, 2010 #3
    We've never talked about centripetal force in class so I doubt that the answer is related to that.

    I do not believe we have to take into consideration relativistic effects.

    Would a formula like F= BIL be more effective? The current seems to be the missing link in all the magnetic field equations given.

    Thank you for your reply.
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