Question

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of $2.0\times {10}^{5}m{s}^{-1}.$ The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton $=1.6\times {10}^{-27}kg.$

Answer 1

$m_p=1.6\times {10}^{-27}kg,$

$v=2\times {10}^{5}m/s,$

$r=4cm=4\times {10}^{-2}m$

Since the proton is not deflected in the combine magnetic and electric field, hence force due to both field must be same.

$i.e.,qE=qvB\Rightarrow E=vB$

Now when the electric field is stopped, then it form a circle due to force of magnetic field.

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

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

$B=0.5\times {10}^{-1}=0.05T$

$E=vB=2\times {10}^5\times 0.05$

$=1\times {10}^{4}N/c$