A particle of mass m and charge q moves with a constant velocity v along the positive x-direction. I

Q: Why does the particle move in a circular path?

The magnetic force $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ is always perpendicular to the velocity. This perpendicular force acts as a centripetal force, causing the particle to move in a circle without changing its speed.

Q: Where is the center of the circular path located?

When the particle enters the field at $x = a$, the magnetic force is entirely in the $y$-direction. Since the force points towards the center of the circle, the center must lie on the line $x = a$.

Q: What happens if $r < b - a$?

If the radius is less than the width of the magnetic field region, the particle will complete a semi-circle inside the field and exit from the same boundary $x = a$ where it entered.

Q: Why is the minimum velocity condition $r > b - a$ and not $r = b - a$?

If $r = b - a$, the particle will exactly graze the boundary $x = b$ and then turn back. To actually *enter* the region $x > b$, the radius must be strictly greater than $b - a$.

Q: Does the sign of the charge affect the condition?

No. A negative charge would experience a force in the opposite direction (downwards instead of upwards), but the geometry of the circular path and the condition to cross the region $r > b - a$ remains exactly the same.

Question

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$-direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$-direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

Options:

(A) $\frac{qbB}{m}$

(B) $\frac{q(b - a)B}{m}$

(C) $\frac{qaB}{m}$

(D) $\frac{q(b + a)B}{m}$

Sigma Percentile · Accepted Answer

### The Invisible Boundary

Imagine you are a charged particle, hurtling through space at a constant velocity $v$. Suddenly, you cross an invisible threshold at $x = a$ and enter a region filled with a uniform magnetic field $\mathbf{B}$. This field is directed straight into the page, along the negative $z$-direction.

What happens next is a beautiful dance dictated by the laws of electromagnetism. The moment you enter this region, you experience a magnetic force. This force is given by the Lorentz force equation:

$$ \mathbf{F} = q(\mathbf{v} 	imes \mathbf{B}) $$

Because your velocity is perfectly perpendicular to the magnetic field, this force acts purely as a centripetal force. It doesn't speed you up or slow you down; it simply turns you, forcing you into a perfect circular path.

### The Geometry of the Curve

The radius of this circular path is a delicate balance between your momentum and the magnetic grip. It is given by the classic formula:

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

Now, let's look at the geometry of your situation. You entered the field exactly at $x = a$, moving perfectly horizontally. This means the magnetic force initially pulls you perfectly vertically. Consequently, the center of your circular path must lie exactly on the entry line, $x = a$.

Since the center of your circle is at $x = a$, the absolute maximum distance you can travel in the forward $x$-direction is simply the center's position plus the radius.

$$ x_{	ext{max}} = a + r $$

### The Three Fates

The magnetic field region has a finite width. It ends at $x = b$. The total width of this magnetic trap is $d = b - a$. Depending on your velocity, and therefore your radius $r$, you face one of three fates:

**Fate 1: The U-Turn**
If your radius is small ($r < b - a$), the magnetic field is too strong for your momentum. You will complete a semi-circle entirely within the field and be spat back out from the exact same boundary you entered, $x = a$.

**Fate 2: The Graze**
If your radius perfectly matches the width of the field ($r = b - a$), you will just manage to touch the far boundary at $x = b$. You will graze it for a fleeting moment before the magnetic force pulls you back around.

**Fate 3: The Breakthrough**
To truly escape the magnetic trap and enter the region $x > b$, your circular path must be large enough to cross the far boundary. Your radius must be strictly greater than the width of the field.

$$ r > b - a $$

### The Final Calculation

We know the condition for breakthrough. Now, we simply substitute our physical formula for the radius into this geometric constraint:

$$ \frac{mv}{qB} > b - a $$

To find the minimum velocity required to achieve this, we rearrange the inequality to solve for $v$:

$$ v > \frac{q(b - a)B}{m} $$

And there we have it! The minimum velocity required to just enter the region beyond the magnetic field is $\frac{q(b - a)B}{m}$.

I know boundary problems can sometimes feel intimidating, but notice how beautifully the physics translates into pure geometry. Whether the charge is positive or negative, the geometric constraint of the circle crossing the boundary remains the universal key to unlocking the solution.

Select Answer:

Visualized Solution

$Initial Setup$

$\text{Lorentz Force & Circular Path}$

$Geometric Constraints$

$Analyzing Trajectories$

$Condition to Cross$

$\text{Conclusion & Extensions}$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Invisible Boundary

The Geometry of the Curve

The Three Fates

The Final Calculation

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