Animated Solution for Physics - Magnetic Effects of Current: A particle of charge $q$ and mass $m$ is moving with a velocity $-v\hat{i}(v\ne 0)$ towards a large screen placed in the $yz$-plane at a distance $d$. If there is a magnetic field $\mathbf{B}=B_0\hat{k}$, the minimum value of $v$ for which the particle will not hit the screen is

Imagine you are observing a tiny charged particle, possessing a mass $m$ and a charge $q$, hurtling through space. It is moving with a velocity $\mathbf{v}=-v\hat{i}$, meaning it is heading straight towards a large screen positioned in the $yz$-plane at a distance $d$.

But this isn't empty space. The entire region is permeated by a uniform magnetic field $\mathbf{B}=B_0\hat{k}$, pointing directly out of the page towards you.

As the particle enters this magnetic field, it doesn't just keep flying straight. It experiences a Lorentz force. We can determine the direction of this force using the cross product $\mathbf{F}=q(\mathbf{v}\times \mathbf{B})$.

Substituting our vectors, we get $\mathbf{F}=q(-v\hat{i}\times B_0\hat{k})$. Using the right-hand rule, the cross product of $-\hat{i}$ and $\hat{k}$ gives us $+\hat{j}$. Therefore, the force is $\mathbf{F}=qv{B}_0\hat{j}$, which points directly upwards.

Because this magnetic force is always perpendicular to the particle's velocity, it acts as a centripetal force. Instead of moving in a straight line, the particle is forced to travel in a circular path.

Our main goal is to ensure that the particle does not hit the screen. For this to happen, the particle must complete its semi-circular turn before it covers the horizontal distance $d$.

Geometrically, the maximum horizontal distance the particle will travel is exactly equal to the radius $R$ of its circular path. Therefore, to avoid a collision, the radius must be less than or equal to the distance to the screen:

If $R$ is exactly equal to $d$, the particle will just graze the screen. If $R>d$, a collision is inevitable.

From our knowledge of electromagnetism, the radius of a charged particle's circular path in a magnetic field is given by equating the magnetic force to the required centripetal force ($qv{B}_0=\frac{m{v}^2}{R}$), which simplifies to:

Now, we substitute this expression for $R$ back into our safety condition:

To find the condition for the velocity $v$, we simply rearrange the inequality:

This inequality tells us that the velocity $v$ must not exceed $\frac{q{B}_{0}d}{m}$. Therefore, the maximum safe velocity is:

A Quick Note on the Question: You might have noticed that the original exam question asks for the minimum value of $v$. This is a known typographical error. Physically, any velocity smaller than our calculated value will result in a tighter, smaller circle, which is perfectly safe. The critical threshold we calculated is actually the maximum allowable velocity before the particle crashes into the screen.