A particle of mass m = 1.6 x 10^-27 kg and charge q = 1.6 x 10^-19 C enters at E in a region of unif

Q: Why doesn't the speed of the particle change in the magnetic field?

The magnetic force is always perpendicular to the velocity vector. Since work done is force times displacement in the direction of force, the magnetic force does zero work. Thus, kinetic energy and speed remain constant.

Q: How do we know the particle completes exactly 1/4th of a circle in Case (a)?

The particle enters at an angle of $45^\circ$ to the boundary. The center of the circular path forms a right isosceles triangle with the entry and exit points. The angle subtended at the center is $90^\circ$, which is exactly $1/4$th of a full $360^\circ$ circle.

Q: Why does the particle exit at the same angle it entered in Case (a)?

The circular path is symmetric with respect to the horizontal axis passing through the center of the circle. Because of this geometric symmetry, the angle of incidence equals the angle of exit.

Q: Why does the particle complete 3/4th of a circle in Case (b)?

When the magnetic field is reversed, the force pushes the particle deeper into the field region. It must complete a full loop minus the $90^\circ$ arc that lies outside the boundary. Thus, it completes $360^\circ - 90^\circ = 270^\circ$, which is $3/4$th of a circle.

Q: Does the mass or charge of the particle affect the angle at which it exits?

No, the exit angle is purely determined by the entry angle and the geometry of the boundary. The mass and charge only affect the radius of the circular path and the time spent in the field, not the geometric angles of the trajectory.

Question

A particle of mass $m = 1.6 	imes 10^{-27}$ kg and charge $q = 1.6 	imes 10^{-19}$ C enters at $E$ in a region of uniform magnetic field of strength $1$ T along the direction shown in figure. The speed of the particle is $10^7$ m/s.
(a) The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at the point $F$. Find the distance $EF$ and the angle $	heta$.
(b) If the direction of the field is along the outward normal to the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at $E$.

Sigma Percentile · Accepted Answer

The journey of a charged particle through a magnetic field is one of the most elegant dances in physics. It is a perfect interplay between linear momentum and a perpendicular force, resulting in a flawless circular trajectory. Let's dive deep into this problem and uncover the beautiful geometry hidden within!

### The Physics of the Dance

When a charged particle enters a uniform magnetic field, it experiences a magnetic force given by the Lorentz force law:
$$ \vec{F} = q(\vec{v} 	imes \vec{B}) $$

Because this force is always perpendicular to the velocity vector, it does no work on the particle. The kinetic energy and speed remain absolutely constant. Instead of speeding up or slowing down, the particle is constantly pulled sideways, forcing it into a circular path. The magnetic force acts as the centripetal force:
$$ \frac{mv^2}{r} = qvB $$

From this, we can easily find the radius of the circular path:
$$ r = \frac{mv}{qB} $$

Let's substitute the given values into our master equation:
$$ r = \frac{(1.6 	imes 10^{-27} 	ext{ kg})(10^7 	ext{ m/s})}{(1.6 	imes 10^{-19} 	ext{ C})(1 	ext{ T})} $$
$$ r = 0.1 	ext{ m} $$

This radius is the key to unlocking the rest of the problem. No matter which way the magnetic field points, the particle will always travel along a circle of radius $0.1 	ext{ m}$.

### Case (a): The Inward Magnetic Field

Imagine the magnetic field pointing directly into the page. As the positively charged particle enters at point $E$ with a velocity angled at $45^\circ$ to the boundary, the right-hand rule tells us that the magnetic force will push it upwards and to the left.

The particle will curve into the magnetic field region, trace out an arc, and eventually exit at point $F$. But how much of the circle does it trace?

If we draw perpendiculars from the velocity vectors at $E$ and $F$, they intersect at the center of the circle, $C$. Because the entry angle is $45^\circ$, the geometry forms a perfect right isosceles triangle between the center $C$ and the boundary. The angle subtended by the arc at the center is exactly $90^\circ$.

This means the particle completes exactly $\frac{1}{4}$th of a full circle!

Because of the beautiful symmetry of the circle, the particle must exit at the exact same angle it entered. Therefore, the exit angle $	heta$ is simply:
$$ 	heta = 45^\circ $$

To find the distance between the entry and exit points, $EF$, we can use the geometry of our right isosceles triangle. The distance $EF$ is the base of the triangle formed by the center and the two points on the boundary:
$$ EF = 2r \cos 45^\circ $$
$$ EF = 2(0.1) \left(\frac{1}{\sqrt{2}}ight) $$
$$ EF = 0.14 	ext{ m} $$

### Case (b): The Outward Magnetic Field

Now, let's flip the script. What happens if the magnetic field points out of the page?

The velocity is the same, but the magnetic field is reversed. According to the right-hand rule, the magnetic force also reverses direction! Instead of being pushed to the left, the particle is now pushed to the right, deeper into the magnetic field region.

It will still travel in a circle of radius $0.1 	ext{ m}$, but its journey will be much longer. It will curve to the right, complete a large loop, and finally strike the boundary again.

How much of the circle does it complete this time? Since the $90^\circ$ arc from Case (a) is now the only part of the circle that lies **outside** the magnetic field, the particle must travel through the remaining portion of the circle. 
$$ 360^\circ - 90^\circ = 270^\circ $$

An angle of $270^\circ$ corresponds to exactly $\frac{3}{4}$th of a full circle!

To find the time spent in the magnetic field, we first need the time period $T$ for a full circle:
$$ T = \frac{2\pi m}{qB} $$

Since the particle only completes $\frac{3}{4}$th of the circle, the time spent is:
$$ t = \frac{3}{4} T = \frac{3}{4} \left(\frac{2\pi m}{qB}ight) = \frac{3\pi m}{2qB} $$

Let's plug in our values and bring it home:
$$ t = \frac{3\pi (1.6 	imes 10^{-27})}{2(1.6 	imes 10^{-19})(1)} $$
$$ t = \frac{3\pi}{2} 	imes 10^{-8} 	ext{ s} $$
$$ t \approx 4.712 	imes 10^{-8} 	ext{ s} $$

And there we have it! By simply understanding the geometry of circular motion and the right-hand rule, we've completely unraveled the particle's journey.

Visualized Solution

$F_{m} = q v B$

$r = \frac{m v}{q B}$

$r = 0.1 m$

$θ = 4 5^{\circ}$

$E F = 0.14 m$

$Path = \frac{3}{4} Circle$

$t = \frac{3}{4} T$

$t = 4.712 \times 1 0^{- 8} s$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Physics of the Dance

Case (a)

Case (b)

Similar Questions

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

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).

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fm​=qvB

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=qBmv​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=0.1 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=45∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EF=0.14 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path=43​ Circle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=43​T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=4.712×10−8 s