Question

4. In an experiment designed to measure the magnitude of a uniform magnetic field, electrons are accelerated from rest through a potential difference of 350 V and then enter a uniform magnetic field that is perpendicular to the velocity vector of the electrons. The electrons travel along a curved path because of the magnetic force exerted on them, and the radius of the path is measured to be 7.5 cm. (a) what is the magnitude of magnetic field? (b) what is the angular speed of electrons?

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Answer 1

Answer:

(a) The accelerating process may be seen as a conversion of potential energy eV into kinetic energy. Since it starts from rest,

21 m eev 2=eV and

v= m ee2eV

= 9.11×10 −31kg2(1.60×10 −19C)(350V)

=1.11×10 7m/s.

(b) r= qBmv

gives

r= eBm eev

= (1.60×10 −19 C)(200×10 −3T)(9.11×10 −31kg)(1.11×10 7m/s)

=3.16×10 −4 m.