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) \(q(b - a)(B / m)\).