Question

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of ${10}^5$ 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.

Answer 1

Given,

$velocity=10m/s,radius=2cm,mass=1.6\times {10}^{27}kg$

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 $2\times {10}^{-2}$

We known

$r=\frac{mv}{qB}$

$B=\frac{mv}{qr}$

$=\frac{1.6\times {10}^{-27}\times {10}^5}{1.6\times {10}^{-19}\times 2\times {10}^{-2}}$

$=0.05T$

Now, putting the value of B in equation 1 then we get,

$E={10}^5\times 0.05=0.5\times {10}^{4}N/C$