Question

A magnetic field set up usign Helmholtz coils is uniform in a smal region and has a magnitude of $0.75T.$ In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coil. Anarrow beam of (single species) charged praticles all accelerated through $15KV$ enters this region in a direction perpendicular to both the axis of the coils and the electroststic field.If the beam remains undeflected when the electrostic field is $9.0{10}^{5}V/m,$ make a simple guess as to what the beam contains.Why is the answer not unique?

Answer 1

Step 1: Balance the kinetic energy.

Given, magnetic field, $B=0.75T$

Accelerating voltage,$V=15KV=15\times {10}^{3}V$

Electrostatic field, $E=9\times {10}^{5}V/m$

Let the charge of the particle be $=e$

Mass of the particle be $=m$
The kinetic energy of the particle,
$eV=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{e}{m}=\frac{v^2}{2V}...\left(i\right)$

Where, $v=$ velocity of the particle Since the beam remains undeflected,forces on the particle due to electric field and magnetic field gets balanced and thus net force becomes zero.

$F_E=F_B$

$e_E=evB$

$\Rightarrow v=\frac{E}{B}...\left(ii\right)$

Step 2: Find $e/m$ ratio.

From equations (i) and (ii), we get

$\Rightarrow \frac{e}{m}=\frac{{\left(\frac{E}{B}\right)}^2}{2V}=\frac{E^2}{2V{B}^2}$

Putting the values,

$\Rightarrow \frac{e}{m}=\frac{(9\times {10}^5{)}^2}{2\times 15\times {10}^3\times (0.75{)}^2}$

$=4.8\times {10}^{7}Ck{g}^{-1}$

This value of specfic charge $e/m$ is equal to the value of deuteron or deuterium ions, however, this is not a unique
answer. Other possible answers are
${\text{He}}^{++},{\text{Li}}^{++},$ etc.

Final Answer: Only $e/m$ ratio is determined.

Value of specific charge $e/m$ is equal to the value of deuteron or deuterium ions,however, this is not a unique answer.