Q27.12 Each of the lettered points at t... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q27.12 (page 682)

Each of the lettered points at the corners of the cube in Fig. Q27.12 represents a positive charge $q$ moving with a velocity of magnitude $v$ in the direction indicated. The region in the figure is in a uniform magnetic field , parallel to the x-axis and directed toward the right. Which charges experience a force due to $\overset{鈬赌}{B}$ ? What is the direction of the force on each charge?

Short Answer

Expert verified

$a$ experiences a force in the negative z direction

b

experiences a force in the positive y direction

d

experiences a force in the negative y direction

$e$ experience a force 45 degrees below the negative z axis in the yz plane

Step by step solution

particle a

Use the right hand rule.

index finger in the y direction. Middle finger in the x direction, thus our thumb points in the negative z direciton. Therefore particle $a$ feels a force in the negative z direction.

particle b

Again we use the right hand rule. Index finger in the z direction. Middle finger in the x direction, thus our thumb points in the y direciton. Therefore particle $b$ feels a force in the y direction.

particle c

Here the particles direction is antiparallel to the magnetic field, so our particle feels no force.

particle d

This is a slightly more difficult application of the right hand rule. We break our force into components to solve this. We have $d_{x}$ and $d_{- z}$ . Note that $d_{x}$ is parallel to the magnetic field and feels no force, so we only have to consider the other component. Index finger in the negative z direction. Middle finger in the x direction, thus our thumb points in the negative y direciton. Therefore particle feels a force in the negative y direction.

particle e

This is similar to particle $d$ , but now we have to consider both components. Here we have a velocity component in the negative z direction and a component in the y direction shown below.

Thus $v_{y}$ feels a force in the negative z direction, and $v_{z}$ feels a force in the negative y direction. Our net force is the sum of these two components. as shown below.