Electric And Magnetic Fields

To solve this problem, we need to find the value of that represents the magnitude of the electric field in the region where the proton moves undeflected under the influence of crossed electric and magnetic fields.

The proton moves at a constant speed of when both the electric field and the magnetic field are present and crossed such that they cancel each other's force on the proton:

1. The force due to the electric field is given by , where is the charge of the proton.

2. The magnetic force when the proton moves in the magnetic field is given by , where is the speed of the proton and is the magnetic field strength.

3. For these forces to cancel each other, the net force must be zero. Therefore, .

Thus, simplifies to:

Next, consider when the electric field is switched off:

1. The proton moves in a circular path, implying the magnetic force provides the centripetal force needed to keep the proton moving in a circle of radius .

2. The centripetal force is .

3. Thus, .

Simplifying gives:

Substitute known values , , , and :

Now substitute and back into the equation for :

Thus, the magnitude of the electric field is , which can be written as . According to the given unit conversion implies:

Therefore, the value of is 0.2, which fits within the expected range of 1,1 as interpreted here.

• Along axial line:
• Along equatorial line:

• Dipole produces axial field at point .
• Dipole produces equatorial field at point .