Respuesta :
a)
[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)
[tex]F_{B_x}=0[/tex]
[tex]F_{B_y}=0[/tex]
b)
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)
[tex]F_{B_x}=0[/tex]
[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (+z axis)
c)
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (+x axis)
[tex]F_{B_x}=3.21\cdot 10^{-3} N[/tex] (+y axis)
[tex]F_{B_y}=3.21\cdot 10^{-3}N[/tex] (-x axis)
Explanation:
a)
The electric force exerted on a charged particle is given by
[tex]F=qE[/tex]
where
q is the charge
E is the electric field
For a positive charge, the direction of the force is the same as the electric field.
In this problem:
[tex]q=+4.9\mu C=+4.9\cdot 10^{-6}C[/tex] is the charge
[tex]E_x=+242 N/C[/tex] is the electric field, along the x-direction
So the electric force (along the x-direction) is:
[tex]F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N[/tex]
towards positive x-direction.
The magnetic force instead is given by
[tex]F=qvB sin \theta[/tex]
where
q is the charge
v is the velocity of the charge
B is the magnetic field
[tex]\theta[/tex] is the angle between the directions of v and B
Here the charge is stationary: this means [tex]v=0[/tex], therefore the magnetic force due to each component of the magnetic field is zero.
b)
In this case, the particle is moving along the +x axis.
The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,
[tex]F_{E_x}=1.19\cdot 10^{-3} N[/tex] (towards positive x-direction)
Concerning the magnetic force, we have to analyze the two different fields:
- [tex]B_x[/tex]: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=0^{\circ}[/tex], so the force due to this field is zero.
[tex]- B_y[/tex]: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, [tex]\theta=90^{\circ}[/tex]. Therefore, [tex]\theta=90^{\circ}[/tex], so the force due to this field is:
[tex]F_{B_y}=qvB_y[/tex]
where:
[tex]q=+4.9\cdot 10^{-6}C[/tex] is the charge
[tex]v=345 m/s[/tex] is the velocity
[tex]B_y = +1.9 T[/tex] is the magnetic field
Substituting,
[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And the direction of this force can be found using the right-hand rule:
- Index finger: direction of the velocity (+x axis)
- Middle finger: direction of the magnetic field (+y axis)
- Thumb: direction of the force (+z axis)
c)
As in part b), the electric force has not change, since it does not depend on the veocity of the particle:
[tex]F_{E_x}=1.19\cdot 10^{-3}N[/tex] (+x axis)
For the field [tex]B_x[/tex], the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is
[tex]F_{B_x}=qvB_x[/tex]
And by substituting,
[tex]F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+x axis)
- Thumb: force (+y axis)
For the field [tex]B_y[/tex], the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is
[tex]F_{B_y}=qvB_y[/tex]
And by substituting,
[tex]F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N[/tex]
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+y axis)
- Thumb: force (-y axis)
