# A particle of charge q and mass m is moving with velocity $\vec{v}$ in the positive x-direction.(a) It is subjected to a uniform magnetic field B directed along negative z-direction. Explain briefly the trajectory it would describe.(b) When the particle is subjected simultaneously to both the magnetic and electric fields directed along the z-axis and y-axis respectively, obtain the condition when the particle will go undeflected.

a) Here the velocity of particle $\vec{v}$ is along the positive x - direction. The magnetic field is along the negative z-direction.

The force experienced by the particle is

$F=q\left [ v\times B \right ]$

So the direction of F is found by applying screw rule or the right-hand thumb rule. It is along the positive y-direction. So the particle makes a circular path, where the radius of the path is

$\frac{mv^2}{r}=qvB$     or

$r=\frac{mv}{qB}$

b)

Here the particle is subjected simultaneously to both magnetic and electric field. So the force experienced is

$F=q(E+v\times B)=F_E+F_B$

Here the condition is Electric field and Magnetic field are perpendicular to each other and the velocity of the particle is perpendicular to both.

Then, we have,

$E=E\hat{j}, B=B\hat{k}, v=v\hat{i}$  (from figure)

So $F_E=qE\hat{j},F_B=q(v\hat{i}\times B\hat{k})=-qB\hat{j}$

Therefore, $F=q(E-vB)\hat{j}$

Thus the electric force and magnetic force are in the opposite direction. So to obtain the condition that the particle will go undeflected adjust the two forces to make magnitude equal. Then total force acting on the particle is zero.

That is

$qE=qvB\: \: \: \: or\: \: \: \: v=\frac{E}{B}$

Which means when the particle have velocity $v=\frac{E}{B}$, the particle will go undeflected.

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