A particle of mass M and positive charge Q, moving with a constant velocity u1 = 4i ms^-1, enters a

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

Because the magnetic force is always perpendicular to the velocity, it acts as a centripetal force, changing the direction of the particle without changing its speed, resulting in circular motion.

Q: How do we know the speed remains constant?

The magnetic force does no work on the particle since it's perpendicular to the displacement. Therefore, the kinetic energy and speed remain constant. You can also verify this by checking the magnitudes of the initial and final velocities.

Q: Why did we use $\tan \theta$ to find the angle?

The velocity vector is tangent to the path. The angle the velocity vector makes with the x-axis is the same as the angle the particle has rotated through the circular arc, assuming it started moving along the x-axis.

Q: What if the charge was negative?

If the charge was negative, the magnetic force would be in the opposite direction for the same velocity and magnetic field. To get the same upward deflection, the magnetic field would have to point in the $+z$ direction.

Q: Why is the time converted to $10^{-2}$ seconds?

The time is given as 10 milliseconds. "Milli" means $10^{-3}$, so $10 \text{ ms} = 10 \times 10^{-3} \text{ s} = 10^{-2} \text{ s}$. It's crucial to use standard SI units for all calculations.

Question

A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\mathbf{u}_1 = 4\hat{\mathbf{i}}	ext{ ms}^{-1}$, enters a region of uniform static magnetic field normal to the $x$-$y$ plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity $\mathbf{u}_2 = 2(\sqrt{3}\hat{\mathbf{i}} + \hat{\mathbf{j}})	ext{ ms}^{-1}$. The correct statement(s) is (are)

Options:

(A) the direction of the magnetic field is $-z$ direction.

(B) the direction of the magnetic field is $+z$ direction

(C) the magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.

(D) the magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracking the journey of a charged particle. It enters a region of magnetic field right at the origin, moving purely along the $x$-axis with an initial velocity of $\mathbf{u}_1 = 4\hat{\mathbf{i}}	ext{ ms}^{-1}$.

When it finally emerges from the other side of this magnetic region, its velocity has shifted to $\mathbf{u}_2 = 2\sqrt{3}\hat{\mathbf{i}} + 2\hat{\mathbf{j}}	ext{ ms}^{-1}$. Notice how the path curves upwards! The presence of a $\hat{\mathbf{j}}$ component in the final velocity tells us that the particle has been deflected in the positive $y$-direction.

### The Right-Hand Rule

Why does it curve upwards? The magnetic force is the invisible hand responsible for this. For a positive charge $Q$ moving initially along the positive $x$-axis to experience a force in the positive $y$-direction, the magnetic field must point directly into the screen.

Using the Lorentz force equation, $\mathbf{F} = Q(\mathbf{v} 	imes \mathbf{B})$, and applying the right-hand rule, $\mathbf{v} 	imes \mathbf{B}$ gives us a force along the positive $y$-axis only if $\mathbf{B}$ is in the negative $z$-direction. Thus, we have immediately deduced the direction of the magnetic field!

### The Geometry of the Path

Now, let's find out exactly how much the particle's path bent. The final velocity vector gives us the perfect clue. The angle $	heta$ it makes with the $x$-axis can be found using basic trigonometry:

$$	an 	heta = \frac{u_{2y}}{u_{2x}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

This means the particle deflected by exactly $30^\circ$, which is equivalent to $\frac{\pi}{6}$ radians.

### The Master Equation

In a uniform magnetic field, a charged particle moves in a circular arc. The angle it sweeps out, $	heta$, is simply its angular velocity, $\omega$, multiplied by the time spent in the field, $t$.

Remember, the angular velocity for a particle in a magnetic field is given by $\omega = \frac{QB}{M}$. Therefore, the angle of deflection is:

$$	heta = \omega t = \left(\frac{QB}{M}ight)t$$

### Final Calculation

Let's plug in what we know. The angle $	heta$ is $\frac{\pi}{6}$. The time $t$ is given as $10	ext{ ms}$, which is $10 	imes 10^{-3}	ext{ s} = 10^{-2}	ext{ s}$.

$$\frac{\pi}{6} = \left(\frac{QB}{M}ight) 	imes 10^{-2}$$

All that's left is to solve for the magnetic field $B$. Rearranging the equation, we move $M$ and the constants to the other side:

$$B = \frac{\pi M}{6Q 	imes 10^{-2}}$$

The $10^{-2}$ in the denominator becomes a $100$ in the numerator.

$$B = \frac{100\pi M}{6Q} = \frac{50\pi M}{3Q}$$

So, the magnitude of the magnetic field is $\frac{50\pi M}{3Q}$, and it points in the $-z$ direction. A beautiful blend of kinematics and electromagnetism!

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Visualized Solution

$Initial and Final Velocities$

$Direction of Magnetic Field$

$Angle of Deflection$

$Angular Velocity$

$Substituting Values$

$Solving for B$

$Final Conclusion$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Right-Hand Rule

The Geometry of the Path

The Master Equation

Final Calculation

Similar Questions

Comprehension Passage

In which case would the particle move in a straight line along the negative direction of Y-axis (i.e. move along $- \overset{y}{^}$ )?

In which case will the particle move in a straight line with constant velocity?

In which case will the particle describe a helical path with axis along the positive z-direction?

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