1) A beam of protons is moving in the + x direction with a speed of 22.2 km/s th

Question

1) A beam of protons is moving in the +x direction with a speed of 22.2 km/s through a region in which the electric field is perpendicular to the magnetic field. The beam is not deflected in this region. If the magnetic field has a magnitude of 0.696 T and points in the +z direction, what is the magnitude of the electric field?

2) What is the direction of the electric field?
+y
-z
-y
+x
-x
+z

3) Would electrons that have the same velocity as the protons be deflected by these fields? If so, in what direction would they be deflected?
Yes, +x
Yes, -y
Yes, -z
No
Yes, -x
Yes, +z
Yes, +y

Explanation / Answer

Given,

v = 22.2 km/s = 2.22 x 10^4 m/s

we know that,

v = E/B => E = v B

E = 2.22 x 10^4 x 0.696 = 1.55 x 10^4 V/m

Hence, E = 1.55 x 10^4 V/m

2)The speed, B field and E field are all perpendicular to each other, so the direction will be:

+y

3)No

charge is same for electron but apposite in polairty, the magnetic force will act in apposite direction and the electric and magnetic force will balance each other so no deflection will occur

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.