Answer:
a) B = 1.01 T , B = 1.22 T
Step-by-step explanation:
For this exercise we will use the equation of the magnetic force
F = q v x B
The bold indicate vectors, we can also work this equation using the modules of the quantities
F = q v B sinθ
The direction of the force is obtained by the rule of the right hand, where the thumb points in the direction of the speed and the extended fingers are in the direction of the magnetic field and the palm of the hand points in the direction of the force, That is for positive charges.
Let's apply the above to our exercise
Case 1. A proton with a positive charge moves in the x direction and the force is in the y direction, and with the right hand rule the magnetic field points in the negative direction of the z axis
The magnitude of the field is
B = F / q v B sin θ
Since v and B are perpendicular the angle is 90 ° therefore sin 90 = 1
B = 2.1 10⁻¹⁶ / (1.6 10⁻¹⁹ 1.3 10³)
B = 1.01 T
Case 2. In this case it is an electron with negative charge
The velocity is in the negative direction of the z axis, the force in the positive direction of the y axis, whereby the field is in the positive direction of the x axis, since the charge is negative
Let's calculate the magnitude of the field
B = 8.20 10⁻¹⁶ / (1.6 10⁻¹⁹ 4.20 10³)
B = 1.22 T
Case c and d. They ask to calculate the force for when an electron moves on the axis –y at 3.70 10³ m / s
F = q v B
F = 1.6 10⁻¹⁹ 3.7 10³ B
F = 5.92 10⁻¹⁶ B
The explicit value of the force depends on the value of the magnetic field.
Using the right hand rule as the charge is negative and the velocity is at –and the force is on the x-axis and the magnetic field is on the z-axis.
We have two possibilities:
- If B is in the positive part of the z axis the force goes in the positive part of the x axis
- If B is in the negative part of the z axis the force is directed to the negative part of the x axis