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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94

Problem 1 Easy Difficulty

$\bullet$ In a 1.25 T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50$\mu \mathrm{C}$ and initially moving northward at 4.75 $\mathrm{km} / \mathrm{s}$ is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

Answer

0.0505$N$

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Video Transcript

for a given problem For about a a particle moving in towards the north is deflected towards the east. That means, um the force is applied in this direction. We can show that by using a right and rule. So there this is Tom pointing this directions or forces in this direction. That means the particle is not negative rather a positive. So particle is supposed to. So this is for body for But maybe we can find the magnitude of the force will be kun, we across b times a sign better. Where better is a call to 90 degrees? We can therefore right kun, we cross b We're, um, using the values given in a problem. Then we get the force to be 0.505 Newton.

We have video lessons for 97.51% of the questions in this textbook
Hugh D. Young

College Physics

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