The region between x = 0 and x = L is filled with uniform steady magnetic field -B0 k. A particle of

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

The magnetic force is always perpendicular to the velocity vector, which provides the necessary centripetal force without changing the particle's speed, resulting in uniform circular motion.

Q: How do we know the radius vector sweeps the same angle as the exit velocity?

The velocity vector is always tangent to the circular path, meaning it is perpendicular to the radius vector. If the tangent rotates by 30 degrees, the radius vector must also rotate by exactly 30 degrees.

Q: Why doesn't the particle cross the expanded magnetic field in part (b)?

The maximum horizontal distance the particle can travel is equal to its orbital radius $R$. Since the new width $2.1 L$ is $1.05 R$, which is greater than $R$, the particle turns back before reaching the other side.

Q: Why is the final velocity in part (b) exactly $-v_0 \hat{\mathbf{i}}$?

Because the particle completes a perfect semicircle, its velocity vector rotates by exactly 180 degrees. Since it entered with $v_0 \hat{\mathbf{i}}$, it must exit with $-v_0 \hat{\mathbf{i}}$.

Q: How is the time spent in the magnetic field calculated for part (b)?

The particle completes a semicircle, which is exactly half of a full circular orbit. Therefore, the time spent is half of the full time period $T = \frac{2\pi m}{qB_0}$.

Question

The region between $x = 0$ and $x = L$ is filled with uniform steady magnetic field $-B_0\hat{\mathbf{k}}$. A particle of mass $m$, positive charge $q$ and velocity $v_0\hat{\mathbf{i}}$ travels along $x$-axis and enters the region of the magnetic field. Neglect the gravity throughout the question.
(a) Find the value of $L$ if the particle emerges from the region of magnetic field with its final velocity at an angle $30^\circ$ to its initial velocity.
(b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now expands upto $2.1 L$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a positively charged particle, full of kinetic energy, cruising along the positive x-axis.

Suddenly, it enters a region filled with a uniform magnetic field.

This field is pointing directly into the page, represented by the vector $-B_0 \hat{\mathbf{k}}$.

Our goal is to understand exactly how this invisible force field bends the particle's trajectory and to calculate the precise dimensions of this region based on the particle's exit angle.

### The Master Equation

The moment the particle enters the magnetic field, it experiences the **Lorentz force**.

Mathematically, this is expressed as $\vec{F}_m = q(\vec{v} 	imes \vec{B})$.

Since the initial velocity is along the x-axis ($v_0 \hat{\mathbf{i}}$) and the magnetic field is into the page ($-B_0 \hat{\mathbf{k}}$), the cross product gives us a force in the positive y-direction.

This force is always perpendicular to the velocity, acting as a pure **centripetal force**.

Consequently, the particle is forced into a perfect circular path.

The radius of this circular path is a classic result in electromagnetism: **$R = \frac{mv_0}{qB_0}$**.

### Geometry of the Exit Point

In part (a) of our problem, we are told that the particle emerges from the magnetic field at $x = L$.

Crucially, its final velocity makes an angle of $30^\circ$ with its initial velocity.

Because the velocity vector is always tangent to the circular path, the angle the velocity vector turns through is exactly equal to the angle swept by the radius vector from the center of the circle.

Therefore, the radius vector also sweeps an angle of $30^\circ$.

### Calculating the Width

Let's look at the right-angled triangle formed by the radius vector at the exit point.

The horizontal distance traveled by the particle, which is the width of the region $L$, is simply the opposite side of this triangle.

Using basic trigonometry, we can write $\sin 30^\circ = \frac{L}{R}$.

Since we know that $\sin 30^\circ = \frac{1}{2}$, this immediately tells us that **$L = \frac{R}{2}$**.

Substituting our expression for the radius $R$, we find the exact width of the region:

**$L = \frac{mv_0}{2qB_0}$**.

### The Expanded Magnetic Field

Now, let's dive into part (b).

The magnetic field region is suddenly expanded to a new width of $2.1 L$.

What happens to our particle now?

Let's substitute our previous finding: the new width is $2.1 	imes \left(\frac{R}{2}ight) = 1.05 R$.

This is a critical revelation!

The maximum horizontal distance the particle can possibly travel in its circular path is exactly equal to its radius $R$.

Since the new width $1.05 R$ is strictly greater than $R$, the particle simply does not have enough "room" to cross the entire region.

### Final Calculation

Because it cannot cross, the particle will continue its circular motion until it turns completely around.

It will complete a perfect semicircle inside the magnetic field and exit from the exact same side it entered ($x = 0$).

Since it has completed a half-circle, its velocity vector has rotated by exactly $180^\circ$.

Therefore, its final velocity is perfectly reversed: **$\vec{v}_f = -v_0 \hat{\mathbf{i}}$**.

Finally, the time spent in the magnetic field is simply half of the full time period of circular motion.

The full time period is $T = \frac{2\pi m}{qB_0}$, so the time spent is **$t = \frac{\pi m}{qB_0}$**.

We have successfully decoded the entire journey of the particle!

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$M a g n e t i c F or ce an d P a t h$

$G eo m e t r y o f t h e E x i t P o in t$

$C a l c u l a t in g W i d t h L$

$F ina l A n s w er f or P a r t (a)$

$E x p an d e d M a g n e t i c F i e l d$

$A na l y z in g t h e N e w P a t h$

$F ina l A n s w er f or P a r t (b)$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

Geometry of the Exit Point

Calculating the Width

The Expanded Magnetic Field

Final Calculation

Similar Questions

A particle of mass $m$ and charge $q$ , moving with velocity $v$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $l$ . Choose the correct choice (s).

The region between $y = 0$ and $y = d$ contains a magnetic field $B = B \hat{k}$ . A particle of mass $m$ and charge $q$ enters the region with a velocity $v = v \hat{i}$ . If $d = \frac{m v}{2 q B}$ , then the acceleration of the charged particle at the point of its emergence at the other side is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Force and Path

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Exit Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Width L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer for Part (a)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanded Magnetic Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the New Path

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer for Part (b)