Question

A magnetic field has a magnitude of 1.20 10-3 T, and an electric field has a magnitude of 5.80 103 N/C. Both fields point in the same direction. A positive 1.8-µC charge moves at a speed of 2.90 106 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Answer 1

0.011 N

Step-by-step 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
`q=1.8 \mu C=1.8 \cdot 10^(-6) C`

The electric force acting on the charge is

`F_E = qE=(1.6 \cdot 10^(-6) C)(5.8 \cdot 10^3 N/C)=9.28 * 10^(-3) N`

and the direction is the same as the electric field, so on the x-axis.

The magnetic force acting on the charge is

`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`

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:

`F= √(F_E^2+F_B^2)= \sqrt{(9.28 \cdot 10^(-3) N)^2+(5.57 \cdot 10^(-3)N)^2} =0.0108 N`

≈ 0.011N

Answer 2

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