Capacitor Exercise (part 2)

Electricity and Magnetism Level pending

A non-relativistic charged particle flies through the electric field of a cylindrical capacitor and gets into a uniform transverse magnetic field with induction $B$ as shown in figure. In the capacitor the particle moves along the arc of a circle, in the magnetic field, along a semicircle of radius $r$ . The potential applied to the capacitor is equal to $V$ , the radii of the electrodes are equal to $a$ and $b$ with $a<b$ . Find the specific charge $\frac{q}{m}$ of the particle. $\frac{q}{m}=\frac{hV^{i}}{cr^{d}B^{e}ln(\frac{b^{f}}{a^{g}})}$ Find $h+i+c+d+e+f+g=?$
This problem is not original

The answer is 9.

1 solution

A Former Brilliant Member
May 1, 2020

Motion through the capacitor :

$\dfrac{qV}{r\ln (\frac{b}{a})}=\dfrac{mu^2}{r}\implies u^2=\dfrac{qV}{m\ln (\frac{b}{a})}$ .

Motion through the magnetic field :

$Bqu =\dfrac{mu^2}{R}\implies u=\dfrac{BqR}{m}$ .

So, $\dfrac{qV}{m\ln (\frac{b}{a})}=\dfrac{B^2q^2R^2}{m^2}\implies \dfrac{q}{m}=\dfrac{V}{B^2R^2\ln (\frac{b}{a})}$ .

Hence the required sum is $1+1+1+2+2+1+1=\boxed 9$ .

@Alak Bhattacharya now it is clearly stated, so please delete the problem

A Former Brilliant Member - 1 year, 1 month ago

Capacitor Exercise (part 2)

The answer is 9.

1 solution

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