The time period of a charged particle undergoing a circular motion in a uniform magnetic field is in

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

The magnetic force is always perpendicular to the velocity vector of the particle. Because work done is the dot product of force and displacement ($W = \vec{F} \cdot d\vec{s}$), a perpendicular force does zero work. With zero work done, the kinetic energy, and therefore the speed, remains perfectly constant.

Q: If a faster particle has a larger radius, why is the time period exactly the same?

It's a beautiful cosmic balancing act! A faster particle does travel quicker, but because $r = \frac{mv}{qB}$, it also travels in a proportionally larger circle. The extra distance it has to cover exactly cancels out its higher speed, meaning it takes the exact same amount of time to complete one lap.

Q: Does this rule hold true for particles moving at the speed of light?

No, it breaks down at relativistic speeds! As a particle approaches the speed of light, its relativistic mass increases. Since $T = \frac{2\pi m}{qB}$, an increase in mass $m$ causes the time period $T$ to increase. This is why standard cyclotrons have a speed limit and must be modified into 'synchrotrons' for ultra-high-speed particles.

Q: What if the particle enters the magnetic field at an angle other than 90 degrees?

If it enters at an angle, its velocity has two components: one perpendicular to the field and one parallel. The perpendicular component causes circular motion, while the parallel component causes it to drift forward, resulting in a helical (corkscrew) path. However, the time taken to complete one circular loop (the pitch period) remains exactly the same: $T = \frac{2\pi m}{qB}$.

Q: How does the sign of the charge affect the time period?

The sign of the charge (positive vs. negative) only determines the direction of the magnetic force (via the right-hand rule), which means it dictates whether the particle revolves clockwise or counter-clockwise. It does not affect the magnitude of the time period at all.

Question

The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

Options:

(A) speed

(B) mass

(C) charge

(D) magnetic induction

Sigma Percentile · Accepted Answer

Imagine you are watching a charged particle, perhaps a tiny proton or an electron, zooming through space. Suddenly, it enters a region where a uniform magnetic field exists, perfectly perpendicular to its path. What happens next is one of the most beautiful dances in physics: the particle is forced into a perfect circular orbit.

But why does it move in a circle, and more importantly, how long does it take to complete one full lap? Let's dive into the mechanics of this motion and uncover a surprising truth about its time period.

### The Master Equation of Circular Motion

For any object to move in a circle, it requires a centripetal force directed towards the center of that circle. In our scenario, this crucial inward pull is provided entirely by the magnetic Lorentz force.

We can write this relationship mathematically by equating the centripetal force to the magnetic force:

$$ \frac{mv^2}{r} = qvB $$

Here, $m$ is the mass of the particle, $v$ is its speed, $r$ is the radius of the circular path, $q$ is the charge, and $B$ is the strength of the magnetic field. This single equation is the foundation of our entire derivation.

### Finding the Radius

By rearranging our master equation, we can easily solve for the radius $r$ of the circular path. Notice that one factor of the speed $v$ cancels out from both sides:

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

This tells us something intuitive: a faster particle (larger $v$) or a heavier particle (larger $m$) will carve out a larger, wider circle. Conversely, a stronger magnetic field (larger $B$) will pull the particle into a tighter, smaller circle.

### The Magic of the Time Period

Now, let's calculate the time period $T$, which is the time it takes for the particle to complete exactly one full revolution. In basic kinematics, time is distance divided by speed. For a circle, the distance is the circumference $2\pi r$:

$$ T = \frac{2\pi r}{v} $$

Let's substitute the expression for the radius $r$ that we just derived into this time period formula:

$$ T = \frac{2\pi}{v} \left(\frac{mv}{qB}\right) $$

Look closely at what happens next. The speed $v$ appears in both the numerator and the denominator. They cancel each other out perfectly!

$$ T = \frac{2\pi m}{qB} $$

### A Profound Conclusion

We are left with a final expression for the time period: **$T = \frac{2\pi m}{qB}$**.

Notice what is missing? There is no $v$ in this equation! This proves a profound and somewhat counter-intuitive result: **the time period of a charged particle in a uniform magnetic field is completely independent of its speed.**

How can this be? If a particle is moving faster, shouldn't it complete the circle quicker? The secret lies in the radius. A faster particle does indeed travel at a higher speed, but it also travels in a proportionally larger circle. The extra distance exactly compensates for the extra speed, resulting in the exact same time period.

This elegant principle is not just a mathematical curiosity; it is the fundamental operating principle behind the **Cyclotron**, a type of particle accelerator that relies on this constant time period to accelerate particles to incredibly high energies using an alternating electric field.

Animated Solution for Physics - Magnetic Effects of Current: The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its

Select Answer:

Visualized Solution

$Circular Motion in B Field$

$Centripetal Force$

$Equating the Forces$

$Radius of the Path$

$Time Period Formula$

$Substituting Radius$

$Final Expression$

$Conclusion$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Master Equation of Circular Motion

Finding the Radius

The Magic of the Time Period

A Profound Conclusion

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