Answer:
0.011 N
Explanation:
Let's take the x,y,z axis such that the electric field E and the magnetic field B are on the x-axis and the velocity of the particle is on the y-axis.
The charge is [tex]q=1.8 \mu C=1.8 \cdot 10^{-6} C[/tex]
The electric force acting on the charge is
[tex]F_E = qE=(1.6 \cdot 10^{-6} C)(5.8 \cdot 10^3 N/C)=9.28 \times 10^{-3} N[/tex]
and the direction is the same as the electric field, so on the x-axis.
The magnetic force acting on the charge is
[tex]F_B = qvB = (1.6 \cdot 10^{-6} C)(2.9 \cdot 10^6 m/s)(1.2 \cdot 10^{-3}T)=5.57 \cdot 10^{-3}N[/tex]
and by using the right hand rule, we find that the direction of the force is on the z-axis.
So, the two forces Fe and Fb are perpendicular to each other. Therefore, the net force acting on the particle is the resultant of the two forces:
[tex]F= \sqrt{F_E^2+F_B^2}= \sqrt{(9.28 \cdot 10^{-3} N)^2+(5.57 \cdot 10^{-3}N)^2} =0.0108 N[/tex]
≈ 0.011N
Given Information:
Magnetic field = B = 1.20x10⁻³ T
Electric field = E = 5.80x10³ N/C
Magnitude of charge = q = 1.8x10⁻⁶ C
speed of charge = v = 2.90x10⁶m/s
Required Information:
Net force on the charge = Fnet = ?
Answer:
Net force on the charge = 0.0121 N
Explanation:
As we know force due to electric field is given by
Fe = qE
Fe = 1.8x10⁻⁶*5.80x10³
Fe = 0.01044 N
As we know force due to magnetic field is given by
Fm = qvB
Fm = 1.8x10⁻⁶*2.90x10⁶*1.20x10⁻³
Fm = 0.006264 N
The net force is given by
Fnet =√(Fe² + Fm²)
Fnet =√((0.01044)² + (0.006264)²)
Fnet = 0.0121 N
Therefore, the net force acting on the charge is 0.0121 N
