A beam of protons is accelerated through a potential difference of 0.740 kV and

Question

A beam of protons is accelerated through a potential difference of 0.740 kV and then enters a uniform magnetic field traveling perpendicular to the field.

Part A What magnitude of field is needed to bend these protons in a circular arc of diameter 1.73 m ? Express your answer in tesla to three significant figures.

Part B

What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

Express your answer in tesla to three significant figures.

Explanation / Answer

Part A

We know the charge of the proton (q), the mass of the proton (m), the electric potential difference (V), the radius and we need to solve for the velocity first using
U = KE
qV = 1/2mv^2
(1.6*10^-19C)(740V) = 1/2(1.67*10^-27kg)v^2 and use simple algebra to solve for v
v = 376558.63 m/s

The magnetic force is equal to the centripetal force Fm = Fc and Fm = qvB and Fc = mv^2/r so
qvB = mv^2/r and one of the v's cancel and we can solve for B

B = mv/qr = (1.67*10^-27kg)(376558.63 m/s) / (1.6*10^-19C)(1.73/2m) = 0.004543 T...........Ans.

Part B

We know the charge of the electron (q), the mass of the electron (m), the electric potential difference (V), the radius and we need to solve for the velocity first using
U = KE
qV = 1/2mv^2
(1.6*10^-19C)(740V) = 1/2(9.*10^-31kg)v^2 and use simple algebra to solve for v
v = 16131329.15 m/s

The magnetic force is equal to the centripetal force Fm = Fc and Fm = qvB and Fc = mv^2/r so
qvB = mv^2/r and one of the v's cancel and we can solve for B

B = mv/qr = (9.1*10^-31kg)(16131329.15 m/s) / (1.6*10^-19C)(1.73/2m) = 0.000106 T...........Ans.

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