An electron enters a region of uniform magnetic
field B at velocity v=10^6 m/s, and travels distance
L=3cm in the field as shown in the figure. The
magnetic field is directed out of the paper with
magnitude B=57mG (1 G = 10^(β4 T)). Due to the
Lorentz force, the electron undergoes a deflection
d above the original line. Calculate d using two
methods. (The charge and mass of electron are
q_e=β1.602 Γ 10-19 C, m_e = 9.109 Γ 10β31 kg)
Method 1: (exact)
1. Since the trajectory of the electron in the magnetic field is an arc (a part of circle), use the Lorentz force and the centripetal acceleration to calculate the radius R of the circle.
2. Calculate the deflection d by geometry: d=R-sqrt(R^2-L^2).
Method 2: (approximate, for d << L)
1. Use the Lorentz force to calculate the acceleration (a) of the electron in the vertical direction. Assume that this acceleration is constant along the path of the electron.
2. Calculate the time (t) for the electron to traverse the distance L, neglecting any deflection (which you should already have found to be much smaller than L).
3. Use the kinematics of constant acceleration in the vertical direction to calculate the deflection.
4. Compare the results from the two methods, and convince yourself that the second method is a good approximation.
Answers: 2
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An electron enters a region of uniform magnetic
field B at velocity v=10^6 m/s, and travels distanc...
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