Magnetism and Matter

Question 1:

Q: A bar magnet of length $L$ and magnetic moment $M$ is placed in a uniform magnetic field $B$ at an angle $\theta$ . Derive the expression for the torque acting on the magnet.

Solution: The torque $\tau$ on a magnetic dipole in a magnetic field is given by:

$$\tau = M B \sin \theta$$

Where:

• $M$ is the magnetic moment,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the magnetic moment and the magnetic field.

Thus, the torque on the bar magnet is $\tau = M B \sin \theta$ .

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Question 2:

Q: Derive the expression for the potential energy of a magnetic dipole in a uniform magnetic field.

Solution: The potential energy $U$ of a magnetic dipole in a uniform magnetic field is given by:

$$U = - \mathbf{M} \cdot \mathbf{B}$$

Where $\mathbf{M}$ is the magnetic moment and $\mathbf{B}$ is the magnetic field.

The dot product can be written as:

$$U = - M B \cos \theta$$

Where:

• $M$ is the magnitude of the magnetic moment,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the magnetic moment and the magnetic field.

Thus, the potential energy of the magnetic dipole is $U = - M B \cos \theta$ .

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Question 3:

Q: A current $I$ flows through a solenoid of length $l$ and $n$ turns per unit length. Derive the expression for the magnetic field inside the solenoid.

Solution: Using Ampère's Law:

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

For a solenoid, the magnetic field inside the solenoid is uniform and parallel to the axis. The integral becomes:

$$B \cdot l = \mu_0 I$$

Where $l$ is the length of the solenoid.

The current $I$ is carried by $n$ turns per unit length. Hence, the total current is $n I l$ , and the magnetic field inside the solenoid is:

$$B = \mu_0 n I$$

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

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Question 4:

Q: A bar magnet is suspended in a uniform magnetic field. Derive the expression for the time period of oscillation of the magnet about its equilibrium position.

Solution: The torque acting on the magnetic dipole is:

$$\tau = M B \sin \theta$$

For small angles $\theta$ , $\sin \theta \approx \theta$ . Thus, the torque is:

$$\tau = M B \theta$$

This is similar to the restoring torque in simple harmonic motion, where the effective moment of inertia of the magnet is $I = I_{\text{magnet}}$ .

The equation of motion is:

$$I \frac{d^2 \theta}{dt^2} = -M B \theta$$

The time period $T$ of oscillation is given by:

$$T = 2\pi \sqrt{\frac{I}{M B}}$$

Thus, the time period of oscillation is $T = 2\pi \sqrt{\frac{I}{M B}}$ .

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Question 5:

Q: The magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is $B$ . Derive an expression for $B$ .

Solution: For a circular current loop, the magnetic field at the center is given by Ampère's Law. The magnetic field at the center of the loop is:

$$B = \frac{\mu_0 I}{2R}$$

Where:

• $\mu_0$ is the permeability of free space,
• $I$ is the current,
• $R$ is the radius of the loop.

Thus, the magnetic field at the center of the loop is $B = \frac{\mu_0 I}{2R}$ .

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Question 6:

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

Solution: Using Ampère’s law, the magnetic field at a distance $r$ from a long straight conductor is:

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

For a circular path around the wire, the length of the path is $2\pi r$ , so:

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

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 7:

Q: A magnetic dipole of moment $M$ is placed in a uniform magnetic field $B$ . Derive an expression for the work done in rotating the dipole from an angle $\theta_1$ to $\theta_2$ .

Solution: The work done $W$ in rotating the dipole is the negative of the change in potential energy:

$$W = - \Delta U$$

The potential energy is:

$$U = - M B \cos \theta$$

The work done in rotating the dipole from $\theta_1$ to $\theta_2$ is:

$$W = - \left[ - M B \cos \theta_2 + M B \cos \theta_1 \right]$$

Thus, the work done is:

$$W = M B (\cos \theta_1 - \cos \theta_2)$$

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Question 8:

Q: A coil of radius $R$ and $n$ turns carries a current $I$ . Derive an expression for the magnetic moment of the coil.

Solution: The magnetic moment $M$ of a coil is given by:

$$M = n I A$$

Where:

• $n$ is the number of turns,
• $I$ is the current,
• $A = \pi R^2$ is the area of the coil.

Thus, the magnetic moment is:

$$M = n I \pi R^2$$

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Question 9:

Q: Derive the expression for the magnetic field at a point on the axis of a circular current loop.

Solution: For a point on the axis of a circular current loop, the magnetic field at a distance $x$ from the center of the loop is given by:

$$B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$$

Where:

• $\mu_0$ is the permeability of free space,
• $I$ is the current in the loop,
• $R$ is the radius of the loop,
• $x$ is the distance of the point from the center of the loop along the axis.

Thus, the magnetic field at the point on the axis is $B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$ .

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Question 10:

Q: A magnet has a magnetic moment $M$ . What is the torque acting on it if it is placed in a magnetic field $B$ at an angle $\theta$ ?

Solution: The torque $\tau$ on a magnetic moment $M$ in a magnetic field $B$ at an angle $\theta$ is given by:

$$\tau = M B \sin \theta$$

Thus, the torque acting on the magnet is $\tau = M B \sin \theta$ .

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Question 11:

Q: Derive an expression for the magnetic field at the center of a circular current loop.

Solution: For a circular loop of current, the magnetic field at the center is given by the Biot-Savart law. The magnetic field at the center of the loop is:

$$B = \frac{\mu_0 I}{2R}$$

Where:

• $\mu_0$ is the permeability of free space,
• $I$ is the current,
• $R$ is the radius of the loop.

Thus, the magnetic field at the center of the loop is $B = \frac{\mu_0 I}{2R}$ .

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Question 12:

Q: Derive the expression for the magnetic field due to a current-carrying solenoid.

Solution: The magnetic field inside a solenoid is given by Ampère’s Law:

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

Where $n$ is the number of turns per unit length and $I$ is the current.

The magnetic field inside the solenoid is uniform, so:

$$B = \mu_0 n I$$

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

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Question 13:

Q: Explain how the magnetic dipole moment of a current loop is related to the area and current.

Solution: The magnetic dipole moment $M$ of a current loop is given by:

$$M = I A$$

Where:

• $I$ is the current flowing in the loop,
• $A$ is the area of the loop.

For a circular loop, the area $A = \pi R^2$ , so the magnetic moment becomes:

$$M = I \pi R^2$$

Thus, the magnetic dipole moment is directly proportional to the current and the area of the loop.

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Question 14:

Q: A current-carrying wire is bent into a semicircular shape. Derive the magnetic field at the center of the semicircular wire.

Solution: The magnetic field at the center of a semicircular current-carrying wire is given by the Biot-Savart law. The magnetic field at the center is:

$$B = \frac{\mu_0 I}{4R}$$

Where:

• $\mu_0$ is the permeability of free space,
• $I$ is the current,
• $R$ is the radius of the semicircle.

Thus, the magnetic field at the center of the semicircular wire is $B = \frac{\mu_0 I}{4R}$ .

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Question 15:

Q: Discuss the Earth's magnetic field and how a magnetic needle aligns itself with the Earth's magnetic field.

Solution: The Earth behaves like a giant magnet with a magnetic field that is similar to that of a bar magnet. The magnetic poles of the Earth are located near the geographic poles, and the magnetic field lines emerge from the south pole and curve around to enter the north pole.

A magnetic needle aligns itself with the Earth's magnetic field because of the torque exerted by the Earth's magnetic field on the magnetic dipole of the needle. The north pole of the magnetic needle points towards the Earth's magnetic south pole, and the south pole of the needle points towards the Earth's magnetic north pole. Thus, the needle aligns itself in the direction of the Earth's magnetic field.

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Question 16:

Q: Show that the magnetic moment of a coil is independent of the angle between the normal to the plane of the coil and the magnetic field.

Solution: The magnetic moment of the coil is given by:

$$M = n I A$$

Where:

• $n$ is the number of turns,
• $I$ is the current,
• $A$ is the area of the coil.

The magnetic moment depends on the current, the number of turns, and the area of the coil. It does not depend on the angle between the normal to the plane of the coil and the magnetic field. This is because the magnetic moment is a property of the coil, and only the torque on the coil changes with the angle, not the magnetic moment itself.

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Question 17:

Q: A current-carrying wire of length $L$ is placed in a magnetic field $B$ . Derive the expression for the force on the wire.

Solution: The force $F$ on a current-carrying wire of length $L$ placed in a magnetic field $B$ is given by:

$$F = I L B \sin \theta$$

Where:

• $I$ is the current in the wire,
• $L$ is the length of the wire,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the magnetic field and the current direction.

Thus, the force on the wire is $F = I L B \sin \theta$ .

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Question 18:

Q: Discuss the behavior of a magnet when placed in a uniform magnetic field.

Solution: When a magnet is placed in a uniform magnetic field, it experiences a torque that tends to align the magnet with the field. The torque on the magnet is given by:

$$\tau = M B \sin \theta$$

Where:

• $M$ is the magnetic moment of the magnet,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the magnetic moment and the magnetic field.

If the magnet is free to rotate, it will align itself with the magnetic field. If it is fixed at an angle, it will experience a restoring torque that tries to bring it into alignment with the field.

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Question 19:

Q: How does the magnetic field behave inside and outside a bar magnet?

Solution: Inside a bar magnet, the magnetic field lines are nearly parallel and strong. The field inside is directed from the south pole to the north pole of the magnet. The magnetic field is uniform in the interior, but as one moves toward the poles, the field weakens.

Outside the bar magnet, the field lines spread out and curve around from the north pole to the south pole. The magnetic field outside the magnet is weaker than inside, and it forms closed loops, connecting the north and south poles.

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Question 20:

Q: A proton and an alpha particle are moving in a magnetic field with the same velocity. Derive the ratio of the radii of their circular paths.

Solution: The magnetic force provides the centripetal force for both the proton and the alpha particle.

For the proton:

$$r_{\text{proton}} = \frac{m_{\text{p}} v}{q_{\text{p}} B}$$

For the alpha particle:

$$r_{\alpha} = \frac{m_{\alpha} v}{q_{\alpha} B}$$

Since the alpha particle has twice the charge and four times the mass of a proton, we can write:

$$\frac{r_{\alpha}}{r_{\text{proton}}} = \frac{m_{\alpha}}{m_{\text{p}}} \cdot \frac{q_{\text{p}}}{q_{\alpha}} = \frac{4}{2} = 2$$

Thus, the radius of the circular path of the alpha particle is twice that of the proton.

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Question 21:

Q: Explain the concept of Earth's magnetic field and its components.

Solution: The Earth’s magnetic field is a dipole field with its magnetic poles located near the geographic poles. The magnetic field lines emerge from the southern hemisphere and enter the northern hemisphere. The components of the Earth’s magnetic field are:

• Magnetic Declination: The angle between the magnetic north and the geographic north.
• Magnetic Inclination: The angle between the magnetic field lines and the horizontal plane.
• Horizontal Component: The component of the magnetic field parallel to the Earth's surface.
• Vertical Component: The component of the magnetic field perpendicular to the Earth's surface.

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Question 22:

Q: Derive an expression for the magnetic field at the center of a uniformly magnetized sphere.

Solution: For a uniformly magnetized sphere, the magnetic field at its center is given by:

$$B = \frac{2\mu_0 M}{3}$$

Where:

• $\mu_0$ is the permeability of free space,
• $M$ is the magnetization of the sphere.

Thus, the magnetic field at the center of a uniformly magnetized sphere is $B = \frac{2\mu_0 M}{3}$ .

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Question 23:

Q: A charged particle moves through a magnetic field. What is the force acting on the particle? Derive an expression for the force in terms of charge, velocity, and magnetic field.

Solution: The force on a charged particle moving in a magnetic field is given by the Lorentz force law:

$$F = q v B \sin \theta$$

Where:

• $q$ is the charge of the particle,
• $v$ is the velocity of the particle,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the velocity and the magnetic field.

Thus, the force on the charged particle is $F = q v B \sin \theta$ .

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Question 24:

Q: Derive the expression for the force on a current-carrying conductor placed in a magnetic field.

Solution: The force on a current-carrying conductor of length $L$ in a magnetic field $B$ is given by:

$$F = I L B \sin \theta$$

Where:

• $I$ is the current,
• $L$ is the length of the conductor,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the current direction and the magnetic field.

Thus, the force on the conductor is $F = I L B \sin \theta$ .

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Question 25:

Q: Explain the concept of magnetization and derive the relation between magnetization and the magnetic field intensity.

Solution: Magnetization $M$ is the magnetic moment per unit volume of a material. The magnetization is related to the magnetic field intensity $H$ by the following relation:

$$M = \chi_m H$$

Where:

• $M$ is the magnetization,
• $\chi_m$ is the magnetic susceptibility,
• $H$ is the magnetic field intensity.

This equation shows that magnetization is proportional to the magnetic field intensity in a material.

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Question 26:

Q: Discuss the principle and working of a moving coil galvanometer.

Solution: A moving coil galvanometer works on the principle that when a current flows through a coil placed in a magnetic field, it experiences a torque that causes it to rotate. The torque is proportional to the current. The deflection of the coil is measured on a scale. A spring or a damping system is used to restore the coil to its zero position when the current stops.

The galvanometer can be calibrated to measure the current by knowing the deflection for a given current.