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 $4.0cm$. 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 undeflected in the combined magnetic and electric field. Hence force due to both the field must be same.

i.e $qE=qvB\Rightarrow R=vB$

Won, when the electric field is stopped, then if forms a circle due to force of magnetic field

We know

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

$\Rightarrow 4\times {10}^2=\frac{1.6\times {10}^{-27}\times 2\times {10}^5}{1.6\times {10}^{-19}\times B}$

$\Rightarrow B=\frac{1.6\times {10}^{-27}\times 2\times {10}^5}{4\times {10}^2\times 1.6\times {10}^{-19}}=0.5\times {10}^{-1}=0.05T$

$E=vB=2\times {10}^5\times 0.05=1\times {10}^{4}N/C$