Since the charge to mass ratio of the electron is known veryaccurately (e/m = 1.

Question

Since the charge to mass ratio of the electron is known veryaccurately (e/m = 1.759 x 1011 C/kg), the experiment youdid in lab can be used to measure the magnitude of a uniformmagnetic field. Electrons are accelerated from rest through apotential difference of 340 V. Thenthese electrons enter a uniform magnetic field that isperpendicular to the their initial directon. Due to the interactionwith the magnetic field, they move in a circular path of radius0.746 cm. What is the magnitude ofthe magnetic field ?

1 mT

Explanation / Answer

the kinetic energy aquired by the electron in movingthru a potential difference V                     mv^2 /2 = qV     m = mass of the electron       v = velocity of theelectron      V = potentialdifference      q = charge on electron                    from the above relaiton we can sovle for the veloctiy of theelectron the force in the electron moving in circularmotion                  mv^2 /r = Bqv                         B= mv /qr                   r = radius of electron                  q = charge on electron                  v = velocity of the electron                 B = magnetic field                  m = mass of the electron from the known values using the above relation we cansolve for B

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