A beam of cathode rays is accelerated from rest through an electric potential difference V… — JAMB UTME

A beam of cathode rays is accelerated from rest through an electric potential difference $V$ before entering a uniform magnetic field $B$ applied perpendicular to the direction of motion, causing the rays to bend into a circular arc of radius $r$ . If the accelerating potential difference is increased to $2V$ and the magnetic field intensity is increased to $2B$ , what is the new radius of curvature of the cathode ray path?

$\frac{r}{\sqrt{2}}$ Answer
B
$r$
C
$\sqrt{2}r$
D
$2r$

Answer

The new radius of curvature is

\frac{r}{\sqrt{2}}

The kinetic energy gained by an electron of mass

m

and charge

e

accelerated through potential difference

V

e V = \frac{1}{2}m v^2

, giving

v = \sqrt{\frac{2eV}{m}}

. When entering a perpendicular magnetic field

B

, centripetal force gives

e v B = \frac{m v^2}{r}

, leading to

r = \frac{m v}{e B} = \frac{1}{B}\sqrt{\frac{2m V}{e}}

. Replacing

V

with

2V

and

B

with

2B

gives

r' = \frac{\sqrt{2}}{2}r = \frac{r}{\sqrt{2}}

Step-by-Step Solution

Relate electron velocity to accelerating potential difference

V

v = \sqrt{\frac{2eV}{m}}

The electrical potential energy lost equals the kinetic energy gained by the cathode ray electrons:

eV = \frac{1}{2}mv^2

Express the radius of curvature

r

in terms of

V

and

B

r = \frac{mv}{eB} = \frac{1}{B}\sqrt{\frac{2mV}{e}}

The magnetic force

evB

provides the necessary centripetal force

\frac{mv^2}{r}

Substitute the scaled values

V' = 2V

and

B' = 2B

into the radius expression

r' = \frac{1}{2B}\sqrt{\frac{2m(2V)}{e}} = \frac{\sqrt{2}}{2}\left(\frac{1}{B}\sqrt{\frac{2mV}{e}}\right) = \frac{r}{\sqrt{2}}

Increasing

V

by a factor of 2 increases

v

\sqrt{2}

, while doubling

B

increases the denominator by 2.

Key Concept

Deflection of cathode rays in magnetic fields and energy conversion of accelerated charges

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