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Find the value of the magnetic field necessary to cause a proton moving at a speed of 2.50 ×10^3 m/s to go into a circular orbit of 15.5-cm radius.

An electron has an energy of 100 eV as it enters a magnetic field of 3.50 × 10−2 T. Find the radius of the orbit.

A velocity selector has a magnetic field of 0.300 T. If a perpendicular electric field of 10,000 V/m is applied, what will be the speed of the particles that will pass through the selector?

  • physics -

    All of these three questions can be solved by applying the rule F = q V B
    which applies when V is perpendicular to V.

    I. e V B = Mp * V^2/R

    You have been given R and V; solve for B.
    e is the proton charge and Mp is the proton mass.

    Try the others yourself

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