In an experiment designed to measure the strength of a uniform magnetic field pr

Question

In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelerated from rest through a potential difference of 207 V. The resulting electron beam travels in a circle with a radius of 3.13 cm. The charge on an electron is 1.60218 Times 10^-19 C and its mass is 9.10939 Times 10^-31 kg. Assuming the magnetic field is perpendicular to the beam, find the magnitude of the magnetic field. Answer in units of T. What is the angular velocity of the electrons?

Explanation / Answer

as the electron is travelling in a circular path

centripetal force = force due to magnetic field

so, mv^2/r = charge x velocity x magnetic field

so, magnetic field= mv/rq = 9.1 x 10^-31 x v/(3.13 x 10^-2 x 1.6 x 10^-19)= 1.817 x 10^-10 x v

now we will calculate v

so 1.6 x 10^-19 x207 = 1/2 x 9.1 x 10^-31 x v^2

so, v= 8.53 x 10^6 m/s

substituting v,

magnetic field B= 1.817 x 10^-10 x 8.53 x 10^6 =1.55 x 10^-3 T

4. angular velocity = v/r= 8.53 x 10^6/3.13 x 10^-2 = 2.72 x 10^8 rad/s

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