1. magnetism

A beam of protons move undeviated through a space having transverse electric and magnetic fields. these fields are mutually perpendicular and their values are 120KV/m and 50mT respectively. if the beam strikes a grounded target then what will be the force exerted by beam on target. given that current is 0.8mA. Assume perfectly inelastic collision.

you know initial velocity as the fields are velocity selector...then what?

2. Originally Posted by ice_syncer
A beam of protons moves undeviated through a space having transverse electric and magnetic fields. these fields are mutually perpendicular and their values are 120KV/m and 50mT respectively. if the beam strikes a grounded target then what will be the force exerted by beam on target. given that current is 0.8mA. Assume perfectly inelastic collision.

you know initial velocity as the fields are velocity selector...then what?

You appear know how to calculate the velocity, $\displaystyle v$, of each electron. Good.

How much momentum does each electron have? It loses that much momentum as it collides with the target.

Use Newton's Second Law in the form: $\displaystyle \vec{F}={{d\vec{p}}\over{dt}}$.

The number of electrons striking the target in a time, $\displaystyle t$, is $\displaystyle n={{I}\over{q_e}}\ t$.

Can you take it from there?

3. no. I reached there myself.

4. just how do you link charge and mass ? dm/dt = dq/dt*dm/dq . I'm stuck here. please continue.

5. Originally Posted by ice_syncer
just how do you link charge and mass ? dm/dt = dq/dt*dm/dq . I'm stuck here. please continue.

The charge of an electron is: -1.602x10^(-19)Coulombs. (Drop the sign.)

The current, I, is 0.8 mA=0.8x10^(-3) Coulombs/second.

The mass of each electron is: 9.11x10^(-31)kg

The mass to charge ratio of an electron is: (9.11x10^(-31)kg)/(1.602x10^(-19)Coulombs). This is dm/dq as you have referred to it.

$\displaystyle F={{dp}\over{dt}}={{d(mv)}\over{dt}}=v{{dm}\over{d t}}=v{{dm}\over{dq}}\cdot{{dq}\over{dt}}=v{{dm}\ov er{dq}}\cdot I$

6. ah ok. thanks