Question

A uniform magnetic field B=3X10^-4 T exist in the +x-direction. a pronton (q=+e) shoots through the field in the +y-direction. find the magnitude and direction of the force on the proton. repeat with the proton replaced by an electron. compare accelleration of those particles. (assueme that the mass of the proton is 2000me)

Answer 1

The force acting on a moving charged particle in a magnetic field is given by :

$F= q\bar{v}\times \bar{B}$

Here,

$q=e$

The value of has not been given, however, the direction is + y direction. Let us therefore denote the velocity as $v\hat{j}$

$B = 3\times 10^{-4} \: \hat{i}$, since it is in the +x direction.

Force acting on the proton will be

$F_{p}= q\bar{v}\times \bar{B}= q(v\hat{j} \times 0.0003\hat{i}) = -0.0003ev\;\hat{k}$

$(\because \hat{j}\times \hat{i}= -\hat{k})$

The force is therefore in the negative z direction.

If the proton is replaced by an electron, q = -e.

$F_{e}= q\bar{v}\times \bar{B}= q(v\hat{j} \times 0.0003\hat{i}) = -0.0003\times (-e)v\;\hat{k}= 0.0003 ev\; \hat {k}$

Hence the magnitude of force acting on the electron and proton is same, but the directions are different.

Force on electron is in +z direction, and force on proton is in -z direction.

But since their masses are different, the accelerations will be different.

$F_{e}=m_{e}a_{e}\: \: \: ,\: \: \: \Rightarrow a_{e}= \frac{F_{e}}{m_{e}}= \frac{0.0003ev}{m_{e}}\hat{k}$

$F_{p}=m_{p}a_{p}\: \: \: ,\: \: \: \Rightarrow a_{p}= \frac{F_{p}}{m_{p}}=\frac{F_{p}}{2000m_{e}}= \frac{-0.0003ev}{2000m_{e}}\hat{k}$

Since proton is 2000 times heavier than electron, the acceleration of proton is 2000 times lesser.

Acceleration of proton is in -z direction, so it curls downwards. Acceleration of electron is in +z direction, so it curls upwards.