A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 105 m/s. The velocity is perpendicular to both the fields. When the electric field is switched off , the proton moves along a circle of radius 2 cm. Find the magnitude of the electric and the magnetic fields. Take the mass of the proton = 1.6 x 10−27 kg.
Given,
velocity=10m/s,radius=2cm,mass=1.6×1027kg
Since the proton is not deflected under the combined action of electric and magnetic field,
So the force applied by the field are equal and opposite,
That, qE=qvB
E=vB.......1
But when the electric field has been stopped proton moves in the circle due to the force of the magnetic field.
The radius of the circle is 2×10−2
We known
r=mvqB
B=mvqr
=1.6×10−27×1051.6×10−19×2×10−2
=0.05T
Now, putting the value of B in equation 1 then we get,
E=105×0.05=0.5×104N/C