Question

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

Short answer

The short answer to the given question is: \(v = q(b - a)(B / m)\), which corresponds to option (b).

Step-by-step solution

1. Identify the Magnetic Force

The magnetic force on a charged particle moving with a velocity in a magnetic field can be given by: \(F = q(\mathbf{v} \times \mathbf{B})\)

2. Calculate the Total Acceleration

As the magnetic force is the only force acting on the charged particle in the region, the total acceleration (a) can be found using Newton's Second Law: \(F = m \times a\)

3. Determine the Time taken to cross Region

Now, we need to find the time taken by the particle to cross the region with a magnetic field (from x = a to x = b) and just enter x > b. We will use the equation of motion, which is: \(x_f = x_0 + v_0t + \frac{1}{2}at^2\)

4. Solve for the minimum velocity

Using the force equation and equation of motion, we can solve for the minimum velocity required to just enter the region x > b.

5. Compare with given expressions

Lastly, we will compare our derived expression for the minimum velocity with the given expressions (a), (b), (c), and (d) and choose the appropriate one. Now, let's perform the calculations: 1. The magnetic force acting on the charged particle is given by: \(\mathbf{F} = \mathrm{q}(\mathrm{v} \times \mathrm{B})\), and it is perpendicular to both velocity and magnetic field. The force magnitude becomes: \(F = \mathrm{q}vb\) 2. Calculate the acceleration of the particle due to magnetic force: \(a = F / m = \mathrm{q}vb / \mathrm{m}\) 3. We are given that the particle is traveling along the x-axis and entering a magnetic field region between x = a and x = b. So, the initial position \(x_0 = a\), and the final position \(x_f = b\). Applying the equation of motion for the x-axis, \(b = a + vt + \frac{1}{2}(\frac{qvb}{m})t^2\) 4. Rearranging the equation and solving for v, we get: \(v = \frac{m}{qb}(b - a)\) 5. Comparing with the given expressions, we can see that our derived expression for the minimum velocity matches expression (b): \(v = q(b - a)(B / m)\) So, the correct answer is (b) \(q(b - a)(B / m)\).

Key concepts

Uniform Magnetic Field

A uniform magnetic field is a region of space where the magnetic force is consistent in both magnitude and direction. This means that any charged particle entering this field experiences the same force throughout its entire motion within the field.
Magnetic fields are generally depicted as lines. These lines show the path along which a positive test charge would move. In a uniform magnetic field, the lines are parallel and evenly spaced, indicating consistent force effects.
This uniform condition makes predicting the behavior of a charged particle more manageable. It allows us to calculate forces and resulting motions with straightforward mathematical expressions.

Charged Particle Motion

When a charged particle enters a magnetic field, it experiences a magnetic force perpendicular to its velocity and the magnetic field direction. This force is calculated as the cross product in vector terms, represented as: \( F = q(\mathbf{v} \times \mathbf{B}) \).
Due to this perpendicular action, the charged particle moves in a circular or helical path. Its direction continuously changes, although its speed remains constant. For the scenario given, as the particle moves along the x-axis into a magnetic field directed along the negative z-direction, it creates a scenario where the path bends but does not slow down.

• The force causes a centripetal acceleration, redirecting the particle rather than changing its speed.
• The particle's kinetic energy remains unchanged, confirming that magnetic forces do no work on the particle since the force is always perpendicular to the direction of motion.

Newton's Second Law

Newton's Second Law is a cornerstone principle of physics, describing how the force acting on an object equates to the product of its mass and acceleration \( F = m \times a \). This principle applies perfectly to charged particle motions in magnetic fields.
In this context, it allows us to express the acceleration of the particle as a result of the magnetic force since no other forces act upon it. Thus, the magnetic force becomes the sole influencer of the particle's motion and path.
When we equate the magnetic force to mass times acceleration, \( qvB = m \times a \), we can easily solve for the particle's acceleration \( a = \frac{qvB}{m} \). This relation helps in understanding how fast the particle's path curves when subjected to magnetic conditions.

Equation of Motion

The equation of motion provides a formula to predict the position of an object over time, considering its initial position, velocity, and acceleration. The general formula is \( x_f = x_0 + v_0t + \frac{1}{2}at^2 \).
Here, it helps us predict how the charged particle moves through the magnetic field region from \( x = a \) to \( x = b \). By substituting known values in, the equation solves for time taken \( t \), allowing us to subsequently determine the required conditions for the particle just to exit the field area.
This motion equation aligns perfectly with defining conditions where the motion is uniform, or whether additional forces, like magnetic ones, alter the predicted trajectory, helping conclude with the correct answer regarding required minimum speed.