Moving Charges And Magnetism

Question 21:

Q: A moving charge $q$ enters a magnetic field at an angle $\theta$ to the field. Derive an expression for the radius of the path it follows.

Solution: When a charged particle enters a magnetic field at an angle $\theta$ to the magnetic field, the magnetic force acts as the centripetal force. The radius $r$ is given by:

1. The magnetic force on the particle:

$$F_B = qvB \sin \theta$$

2. The centripetal force required for circular motion:

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

Equating the magnetic force and the centripetal force:

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

Solving for the radius $r$ :

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

Thus, the radius of the circular path is $r = \frac{mv}{qB \sin \theta}$ .

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Question 22:

Q: A proton is moving with velocity $v$ in a magnetic field $B$ . Derive the expression for the frequency of revolution of the proton.

Solution: The proton moves in a circular path due to the magnetic force, which provides the centripetal force. The frequency $f$ is the inverse of the time period $T$ , and the time period is the time it takes for the proton to complete one revolution.

1. The magnetic force is:

$$F_B = qvB$$

Where $q$ is the charge of the proton, $v$ is the velocity, and $B$ is the magnetic field strength.

2. The centripetal force is:

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

Equating the magnetic force and the centripetal force:

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

Solving for the radius $r$ :

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

The time period $T$ is the time taken to complete one revolution, so:

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

The frequency $f$ is the inverse of the time period:

$$f = \frac{1}{T} = \frac{qB}{2\pi m}$$

Thus, the frequency of revolution is $f = \frac{qB}{2\pi m}$ .

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Question 23:

Q: A current $I$ flows through a long straight conductor. Derive the expression for the magnetic field at a point at distance $r$ from the wire.

Solution: Using Ampère's law, the magnetic field at a distance $r$ from a long straight conductor carrying current $I$ is given by:

1. Ampère's Law states:

$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$$

2. The magnetic field around the wire is tangential to a circle centered on the wire. The length of the circular path is $2\pi r$ , so the integral becomes:

$$B \times 2\pi r = \mu_0 I$$

3. Solving for $B$ :

$$B = \frac{\mu_0 I}{2\pi r}$$

Thus, the magnetic field at a distance $r$ from the wire is $B = \frac{\mu_0 I}{2\pi r}$ .

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Question 24:

Q: A charged particle of mass $m$ and charge $q$ moves with a velocity $v$ in a uniform magnetic field $B$ . Derive the expression for the kinetic energy of the particle after it completes one revolution in the magnetic field.

Solution: When a charged particle moves in a magnetic field, it follows a circular path. The kinetic energy of the particle is given by:

1. The magnetic force provides the centripetal force for circular motion:

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

Solving for $r$ :

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

2. The work done by the magnetic field is zero, as the force is perpendicular to the velocity. Therefore, the kinetic energy of the particle remains constant.

3. The kinetic energy $K$ is:

$$K = \frac{1}{2}mv^2$$

Thus, the kinetic energy of the particle after one revolution is $K = \frac{1}{2}mv^2$ , which remains constant throughout the motion.

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Question 25:

Q: A charged particle enters a uniform magnetic field at an angle $\theta$ . Derive an expression for the time period of the particle's motion.

Solution: When a charged particle moves in a magnetic field at an angle $\theta$ , the force acting on the particle is $F = qvB \sin \theta$ , and this force causes the particle to move in a helical path. The time period $T$ is the time taken to complete one full revolution.

The motion of the particle can be decomposed into two components:

• A circular motion due to the perpendicular component of velocity.
• A linear motion along the direction of the magnetic field due to the parallel component of velocity.

The magnetic force provides the centripetal force for the circular motion:

$$qv_\perp B = \frac{mv_\perp^2}{r}$$

Where $v_\perp = v \sin \theta$ . Solving for the radius $r$ :

$$r = \frac{mv_\perp}{qB} = \frac{mv \sin \theta}{qB}$$

The time period $T$ is the time taken to complete one revolution:

$$T = \frac{2\pi r}{v_\perp} = \frac{2\pi \times \frac{mv \sin \theta}{qB}}{v \sin \theta}$$

Simplifying:

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

Thus, the time period is $T = \frac{2\pi m}{qB \sin \theta}$ .

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Question 26:

Q: Derive the expression for the torque experienced by a current loop in a uniform magnetic field.

Solution: The torque $\tau$ on a current loop in a magnetic field $B$ is given by the formula:

$$\tau = \mu B \sin \theta$$

Where:

• $\mu$ is the magnetic dipole moment of the loop,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the magnetic moment and the magnetic field.

The magnetic dipole moment $\mu$ for a current loop of area $A$ and current $I$ is:

$$\mu = I A$$

Thus, the torque on the loop is:

$$\tau = I A B \sin \theta$$

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Question 27:

Q: A proton moves with velocity $v$ in a uniform magnetic field $B$ perpendicular to the velocity. Derive the expression for the radius of the circular path.

Solution: The magnetic force on the proton acts as the centripetal force for its circular motion. The magnetic force is:

$$F_B = qvB$$

Where $q$ is the charge of the proton and $v$ is the velocity.

The centripetal force required for circular motion is:

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

Equating the magnetic force and centripetal force:

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

Solving for the radius $r$ :

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

Thus, the radius of the circular path is $r = \frac{mv}{qB}$ .

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Question 28:

Q: A solenoid has $n$ turns per unit length. Derive the expression for the magnetic field inside the solenoid.

Solution: For a solenoid, the magnetic field inside the solenoid is uniform and directed along its axis. Using Ampère’s Law:

1. Ampère’s Law is:

$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$$

2. The magnetic field is constant and along the axis of the solenoid. The line integral around the solenoid is:

$$B \times (2\pi r) = \mu_0 n I$$

Where:

• $n$ is the number of turns per unit length,
• $I$ is the current.

Solving for $B$ :

$$B = \mu_0 n I$$

Thus, the magnetic field inside the solenoid is $B = \mu_0 n I$ .

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Question 29:

Q: A charged particle is moving in a uniform magnetic field with velocity $v$ . What is the condition for the particle to move in a circular path?

Solution: For the charged particle to move in a circular path, the magnetic force must provide the centripetal force.

1. The magnetic force on the particle is:

$$F_B = qvB \sin \theta$$

Where $\theta$ is the angle between the velocity and magnetic field.

2. The centripetal force required for circular motion is:

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

Equating the magnetic force and centripetal force:

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

This simplifies to:

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

For the particle to move in a perfect circular path, the angle $\theta$ must be 90° (perpendicular to the magnetic field), so:

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

Thus, for circular motion, $\theta = 90^\circ$ is the condition.

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Question 30:

Q: Explain the phenomenon of magnetic deflection in a cathode ray tube (CRT) and derive the expression for the radius of the circular path of an electron in the magnetic field.

Solution: In a cathode ray tube (CRT), electrons are accelerated and directed through a magnetic field. The magnetic field causes the electrons to follow a circular path due to the Lorentz force acting on them.

The magnetic force provides the centripetal force for the motion of the electron:

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

Solving for $r$ :

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

Thus, the radius of the circular path of an electron in the magnetic field is $r = \frac{mv}{qB}$ .