2. The figure represents the path of an electron that enters with a kinetic ener

Question

2. The figure represents the path of an electron that enters with a kinetic energy KE = 30 keV into a block of material in a region of uniform magnetic field B. After following a half-circular path, the electron exits at a distance of L = 30 cm. fromthe entrance point. particle (a) What is the magnitude and direction of the magnetic field in the material? (b) If the electron has the original kinetic energy KE but exits at a distance of JL from the entrance point, then what would be the magnetic field strength? (c) If the electron exits at the original exit point L but travels with half the kinetic energy 1KE, then what would be the magnetic field strength? (d) If instead of an electron in the original problem, the particle had been a proton, what would have been your answer for the first part? 3. A cosmic-ray proton approaches Earth directly toward the ground somewhere in the Arctic, where the Earth's magnetic field is 0.5 G 80° below the horizontal northward direction. The proton moves in a helical path (due to the magnetic force) with a

Explanation / Answer

Neglecting relativity

0.5m*v*v=30000*1.6*10(-19)

We got v=1.027*10(8) m/s

Radius of circular orbit=L/2=0.15 m

Using

qvB=mv*v/r

We got B=mv/qr=3.9*10(-3) T

(b)if r=0.075

B=7.8*10(-3) T

(c) if ke is halved

V will become 1/sqrt(2) times so B=2.75*10(-3) T

(d) if proton

v=2.39*10(6) m/s

B=0.167 T

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