Physics, 31.07.2019 11:30, meababy2009ow9ewa

Aparticle has a charge of q = +4.8 μc and is located at the origin. as the drawing shows, an electric field of ex = +240 n/c exists along the +x axis. a magnetic field also exists, and its x and y components are bx = +1.6 t and by = +1.7 t. calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.

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Solution
The charge is $q=+4.8 \mu C=+4.8 \cdot 10^{-6} C$

(a) The particle is stationary.
- The magnitude of the force exerted by the electric field is given by
$F_E = qE=(+4.8 \cdot 10^{-6} C)(240 N/C)=1.15 \cdot 10^{-3} N$
the direction of this force is the same as the direction of the field, so along the positive x-axis.
- The Lorentz force, which is the force exerted by a magnetic field B on a particle moving with speed v, is
$F=qvB sin \theta$
where $\theta$ is the angle between the direction of v and B. In this part of the problem the particle is stationary, so v=0 and the two fields Bx and By do not exert any force on the particle, so
$F_{Bx}=F_{By}=0$

(part b) the particle is moving with velocity v_x = 345 m/s

along positive x-axis
- The intensity of the electric force does not depend on the velocity of the particle, therefore it is the same as before:
$F=qE=1.15 \cdot 10^{-3} N$
- The field Bx is parallel to the direction of motion of the particle (x-axis). Therefore, the angle between v and B is $0^{\circ}$ , and so the force exerted by this field is still zero:
$F_{Bx}=0$
- Instead, the field By (y-axis) is perpendicular to the direction of motion of the particle (x-axis), so $\theta = 90^{\circ}$ and $\sin \theta = 1$ . Therefore, the intensity of the force is given by
$F_{By}=qvB_y = (4.8 \cdot 10^{-6} C)(345 m/s)(1.7 T)=2.82 \cdot 10^{-3} N$
Using the thumb rule, we can find the direction of this force:
v-index finger - positive x-axis
B-middle finger - positive y-axis
F-thumb -positive z-axis --> this is the direction of $F_{By}$

(part 3) the particle is moving along positive z-direction with v_z = 345 m/s

- The electric field does not change because it does not depend on the velocity, so
$F_E = 1.15 \cdot 10^{-3} N$
still along positive x-direction.
- Bx: the angle between v and Bx is $90^{\circ}$ , so $sin \theta=1$ . Therefore the intensity of the force is
$F_{Bx}=qvB_x = 2.66 \cdot 10^{-3} N$
Applying the thumb rule, we find that the direction of this force is along positive y-axis.
- By: the angle between v and By is $90^{\circ}$ , so $sin \theta=1$ . Therefore the intensity of the force is
$F_{By}=qvB_y = 2.82 \cdot 10^{-3} N$
And by using again the thumb rule, we find that the direction of this force is along negative x-axis.

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Aparticle has a charge of q = +4.8 μc and is located at the origin. as the drawing shows, an electri...