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drek231 [11]
3 years ago
12

An electron is traveling in the positive x direction. A uniform electric field is present and oriented in the negative z directi

on. If a uniform magnetic field with the appropriate magnitude and direction is simultaneously generated in the region of interest, the net force on the electron can be made to have a magnitude of zero. What must the direction of the magnetic field be?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer:

the fingers extended points in the negative direction of the y-axis

Explanation:

For this exercise we use the relationship of the force with the electric field

          F = q E

electron charge is q = -e

          F = - eE

therefore the forces are in the opposite direction to the electric field direction, which indicates that it goes towards the negative side of the Z axis.

Consequently the electric force is in the positive direction of the z axis

us the direction of the magnetic field so that it cancels out the electric forces.

For the direction of the magnetic force we use the right hand rule. The thumb points in the direction of the speed, the fingers extended in the direction of the magnetic field and the palm in the direction of the Force if the charge is positive, if the charge is negative it is in the opposite direction

In this case we want the magnetic force to be directed in the negative direction of the z axis.

The charge moves in the positive direction of the x-axis

the palm must point in the negative directional of the Z axis, ie the negative charge

therefore the fingers extended points in the negative direction of the y-axis

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3 years ago
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Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) H
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Answer:

(1) 42.94 m

(2) 16.02^\circ

Explanation:

Let us first draw a figure, for the given question as below:

In the figure, we assume that the person starts walking from point A to travel 11 m exactly 24^\circ south of west to point B and from there, it walks 21 m exactly 39^\circ west of north to reach point C.

Let us first write the two displacements in the vector form:

\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m

Part (1):

the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m

Part (2):

As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ

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3 years ago
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In the specific case of the displacement, it is defined as the distance in a straight line between the initial and final position (is a vector magnitude).

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Displacement s the the change in position relative to an initial location.

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