A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the

Q: Why does the particle move in a straight line when both fields are present?

When both fields are present and perpendicular to each other, they exert forces in opposite directions. If their magnitudes are equal ($qE = qvB$), the net force becomes zero, and the particle moves unaccelerated in a straight line.

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

No, it doesn't! Both the electric force ($qE$) and magnetic force ($qvB$) are proportional to the charge $q$. When you equate them, the charge $q$ cancels out, leaving $v = E/B$. This works for both positive and negative charges.

Q: Why do we use $R = mv/qB$ for the circular path?

In a uniform magnetic field, the magnetic force acts perpendicular to the velocity, providing the centripetal force required for circular motion. Equating $qvB = mv^2/R$ gives us this radius formula.

Q: What if the electric field was parallel to the magnetic field?

If the fields were parallel, the electric field would accelerate the particle along the field lines, while the magnetic field would make it spiral. The resulting path would be a helix with increasing pitch, not a straight line.

Q: How do we handle the units in this calculation?

Always convert all given values to standard SI units before calculating. Here, the radius is given in cm, so it must be converted to meters ($0.5 \times 10^{-2}$ m) to ensure the final mass is in kilograms.

Question

A particle having the same charge as of electron moves in a circular path of radius $0.5	ext{ cm}$ under the influence of a magnetic field of $0.5	ext{ T}$. If an electric field of $100	ext{ V/m}$ makes it to move in a straight path, then the mass of the particle is (Take, charge of electron $= 1.6 	imes 10^{-19}	ext{ C}$)

Options:

(A) $1.6 	imes 10^{-19}	ext{ kg}$

(B) $1.6 	imes 10^{-27}	ext{ kg}$

(C) $9.1 	imes 10^{-31}	ext{ kg}$

(D) $2.0 	imes 10^{-24}	ext{ kg}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are observing a charged particle in a laboratory. Initially, it is moving through a region containing only a uniform magnetic field. As expected, the magnetic force acts perpendicular to its velocity, causing it to trace a perfect circular path.

This is our first scenario. The magnetic force provides the necessary centripetal force to keep the particle in orbit.

But then, we introduce a twist. We turn on an electric field in the same region. Suddenly, the particle stops curving and starts moving in a perfectly straight line! This means the new electric force is exactly balancing the magnetic force.

### The Master Equations
Let's break down the physics mathematically. For the first case, the radius of the circular path is governed by the balance of magnetic and centripetal forces:
$$qvB = \frac{mv^2}{R}$$

Rearranging this, we get the classic formula for the radius:
$$R = \frac{mv}{qB}$$

Now, for the second case, the particle moves straight. This implies the net force is zero. The electric force must be equal and opposite to the magnetic force:
$$qE = qvB$$

This gives us the velocity of the particle, a concept famously known as the **velocity selector**:
$$v = \frac{E}{B}$$

### Combining the Concepts
We have two powerful equations. Let's substitute the velocity from the second equation into our radius formula:
$$R = \frac{m}{qB} \left(\frac{E}{B}ight)$$

Simplifying this, we get a direct expression for the mass of the particle:
$$m = \frac{qB^2 R}{E}$$

### Final Calculation
Now, it's just a matter of plugging in the given values. But beware of the units! The radius is given in centimeters, so we must convert it to meters: $R = 0.5 	imes 10^{-2} 	ext{ m}$.

Substituting the values:
$$m = \frac{(1.6 	imes 10^{-19}) 	imes (0.5)^2 	imes (0.5 	imes 10^{-2})}{100}$$

Let's simplify the numerator:
$$m = \frac{1.6 	imes 10^{-19} 	imes 0.25 	imes 0.5 	imes 10^{-2}}{100}$$
$$m = \frac{0.2 	imes 10^{-21}}{100}$$

Finally, dividing by 100 gives us the mass:
$$m = 2.0 	imes 10^{-24} 	ext{ kg}$$

This elegant problem beautifully combines circular motion and the velocity selector principle, a true favorite in competitive exams!

Select Answer:

Visualized Solution

Analyzing the Two Scenarios

Radius of Circular Path

Velocity Selector Condition

Expression for Mass

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup Imagine you are observing a charged particle in a laboratory

The Master Equations Let's break down the physics mathematically

Combining the Concepts We have two powerful equations

Final Calculation Now, it's just a matter of plugging in the given values

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