Question

A tightly wound long solenoid has ‘n’ turns per unit length, a radius ‘r’ and carries a current I. A particle having charge ‘q’ and mass ‘m’ is projected from a point on the axis in a direction perpendicular to the axis. The maximum speed of the particle for which the particle does not strike the solenoid is

• A. \(\frac{\mu_onlqr}{2m}\)
• B. \(\frac{\mu_onlqr}{m}\)
• C. \(\frac{\mu_onlqr}{4m}\)
• D. \(\frac{\mu_onlqr}{8m}\)

Answer 1

The Correct Option is B

1. Magnetic Field inside a Long Solenoid:

The magnetic field inside a tightly wound long solenoid with ‘n’ turns per unit length carrying a current I is given by:

$B = \mu_0 n I$

where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current.

2. Force on a Charged Particle in a Magnetic Field:

A particle with charge ‘q’ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a magnetic force $\vec{F}$ given by the Lorentz force formula:

$\vec{F} = q (\vec{v} \times \vec{B})$

Since the particle is projected perpendicular to the axis of the solenoid (which is the direction of the magnetic field), the velocity $\vec{v}$ is perpendicular to $\vec{B}$. Therefore, the magnitude of the force is:

$F = q v B \sin(90^\circ) = q v B$

3. Circular Motion of the Charged Particle:

The magnetic force acts perpendicular to the velocity of the particle, causing it to move in a circular path in a plane perpendicular to the magnetic field. The magnetic force provides the necessary centripetal force for this circular motion.

Centripetal force $F_c = \frac{m v^2}{R}$, where $m$ is the mass of the particle and $R$ is the radius of the circular path.

Equating the magnetic force to the centripetal force:

$q v B = \frac{m v^2}{R}$

4. Radius of the Circular Path:

From the above equation, we can solve for the radius $R$ of the circular path:

$R = \frac{m v}{q B}$

5. Condition for the Particle Not to Strike the Solenoid:

For the particle not to strike the solenoid, the radius of the circular path must be less than or equal to the radius of the solenoid ‘r’. For the maximum speed, we consider the case where the radius of the circular path is exactly equal to the radius of the solenoid:

$R = r$

$r = \frac{m v_{max}}{q B}$

6. Substitute the Magnetic Field and Solve for Maximum Speed:

Substitute the expression for the magnetic field inside the solenoid, $B = \mu_0 n I$, into the equation for the radius:

$r = \frac{m v_{max}}{q (\mu_0 n I)}$

Now, solve for the maximum speed $v_{max}$:

$v_{max} = \frac{r q (\mu_0 n I)}{m}$

$v_{max} = \frac{\mu_0 n I q r}{m}$

7. Compare with the Given Options:

Comparing the derived expression for $v_{max}$ with the given options, we find that it matches option 1:

$\frac{\mu_0 n l q r}{m}$

Final Answer: The final answer is ${\frac{\mu_0 n l q r}{m}}$

Answer 2

For a tightly wound long solenoid with $n$ turns per unit length, a radius $r$, and carrying a current $I$, a charged particle of charge $q$ and mass $m$ is projected from a point on the axis in a direction perpendicular to the axis. The magnetic field inside the solenoid is given by:

$B = \mu_0 n I$

The force on the charged particle due to this magnetic field is given by the Lorentz force:

$F = q v B$

where $v$ is the velocity of the particle.

To avoid striking the solenoid, the magnetic force must provide the centripetal force for the particle's motion. The maximum speed for which the particle does not strike the solenoid is given by the condition where the force acting on the particle balances the required centripetal force.

The centripetal force is given by:

$F_c = \frac{m v^2}{r}$

Equating the magnetic force and the centripetal force:

$q v B = \frac{m v^2}{r}$

Substitute $B = \mu_0 n I$:

$q v (\mu_0 n I) = \frac{m v^2}{r}$

Solving for $v$, we get:

$v = \frac{\mu_0 n I r}{m}$

Thus, the maximum speed of the particle for which it does not strike the solenoid is:

$v = \frac{\mu_0 n I r}{m}$

So, the correct answer is \(\frac{\mu_0 n I r}{m}\).