Question

A proton of velocity 10 m/s is projected at right angles to a uniform magnetic
induction field of 0.15 T.
(a) How much is the particle path deflected from a straight line after it has traversed a
distance of 1 cm?
(b) How long does it take a proton to traverse a 90° arc?​

Answer 1

Given that,

Velocity of proton $v=10^{7}\ m/s$

Magnetic field = 0.15 T

Distance = 1 cm

(a). We need to calculate the radius

Using formula of electromagnetic field

$F=\dfrac{mv^2}{r}$

$evB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{eB}$

Where, m = mass of proton

e = charge of proton

B = magnetic field

v = speed of proton

Put the value into the formula

$r=\dfrac{1.67\times10^{-27}\times10^7}{1.6\times10^{-19}\times0.15}$

$r=0.696\ m$

$r=69.6\ cm$

We need to calculate the deflection angle

Using formula of deflection angle

$\theta=\dfrac{s}{r}$

Put the value into the formula

$\theta=\dfrac{1}{69.6}$

$\theta=0.014\ rad$

(b). If $\theta=90^{\circ}$

We need to calculate the time

Using formula of time

$t=\dfrac{s}{v}$

$t=\dfrac{\theta\times r}{v}$

Put the value into the formula

$t=\dfrac{\dfrac{\pi}{2}\times0.696}{10^{7}}$

$t=1.09\times10^{-7}\ sec$

Hence, (a). The deflection angle is 0.014 rad

(b). The time is $1.09\times10^{-7}\ sec$