Q: Consider a beam of electrons moving at 2.07 x 10 6 m/s. When a uniform magnet

Question

Q: Consider a beam of electrons moving at 2.07 x 106m/s. When a uniform magnetic field is applied perpendicular to thebeam, the electrons travel in a circular path of radius 0.425 m.Find the magnitude of the magnetic field. Give the answer inmicro-Tesla (T) and to one decimalplace. (micro = = 10-6) Q: Consider a beam of electrons moving at 2.07 x 106m/s. When a uniform magnetic field is applied perpendicular to thebeam, the electrons travel in a circular path of radius 0.425 m.Find the magnitude of the magnetic field. Give the answer inmicro-Tesla (T) and to one decimalplace. (micro = = 10-6)

Explanation / Answer

The force applied to each electron is given by F = qvB. Since this force is making the electron travel in a circle, it is acentripetal force. We can also write the centripetal force as F =mv^2/r Now we equate these to get: qvB = mv^2/r Solving for B gives: B = mv/(qr) Now we just need to plug in the following values: m = mass of an electron = 9.11*10^-31 kg v = velocity = 2.07*10^6 m/s q = charge of an electron = 1.60*10^-19 C r = radius of circle = .425 m Plugging these in gives: ' B = (9.11*10^-31*2.07*10^6)/(1.6*10^-19*.425) B = 2.8*10^-5 T B = 28 T

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