Part A Problem 24.23 What is the magnitude of the magnetic force in figure (a)?

Question

Part A Problem 24.23 What is the magnitude of the magnetic force in figure (a)? Express your answer with the appropriate units An electron moves with a speed of 2.0x10m/s in the directions shown in (Figure 1) below. A 0.60 T magnetic field points in the positive x-direction FBValue Units Submit My Answers Give Up Part B What is the direction of the magnetic force in the figure (a)? r- z direction O +z direction ..-z direction yz direction Figure 1 of 1 Submit My Answers Give Up Correct Part C What is the magnitude of the magnetic force in figure (b)? Express your answer with the appropriate units FBValue Units

Explanation / Answer

Given speed v = 2 * 10^7 m/s

magnetic field B = 0.6 T

charge of electron q = - 1.6 * 10^-19

the magnetic force F = q * v * B * sin(theta)

here   theta = 90

FB = 1.6 * 10^-19 * 0.6 * 2 * 10^7 * sin(90)

FB = 1.92 * 10^-12 N

part B)

the direction is given by right hand thumb rule for positive charge

For negative charge it is opposite so

-Z direction

part c)

Here v is in Y-Z plane so the angle between 'v' and 'B' is 90 deg so

FB = 1.6 * 10^-19 * 0.6 * 2 * 10^7 * sin(90)

FB = 1.92 * 10^-12 N

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