Magnetism And Matter Multiple Choice Questions And Answers

Magnetism And Matter

Question 1. A bar magnet oscillates in the earth’s magnetic field with a period T. What happens to its period and motion if its mass is quadrupled?

1. Its motion remains simple harmonic with time period = T/2.
2. Its motion remains simple harmonic with time period = 2T₁.
3. Its motion remains simple harmonic with time period = 4T.
4. Its motion remains simple harmonic and the time period remains nearly constant.

Answer: 2. Its motion remains simple harmonic with time period = 2T₁.

The time period of a bar magnet oscillating in a magnetic field B is

⇒ \(T=2 \pi \sqrt{\frac{I}{m B}}\)

where I = moment of inertia about the axis of oscillation.

⇒ But I = \(M k^2 \Rightarrow T \propto \sqrt{\text { mass }}\)

⇒ \(\frac{T_2}{T_1}=\sqrt{\frac{4 M}{M}}=\sqrt{4}=2, \text { so } T_2=2 T_1\)

∴ Its motion will be SHM with time period 2T₁.

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Question 2. A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that \(\vec{B}\) is in the plane of the coil. If due to a current i in the triangle a torque τ acts on it, the side l of the triangle is

1. \(\frac{2}{\sqrt{3}}\left(\frac{\tau}{B i}\right)\)
2. \(2\left(\frac{\tau}{\sqrt{3} B i}\right)^{1 / 2}\)
3. \(\frac{2}{\sqrt{3}}\left(\frac{\tau}{B i}\right)^{1 / 2}\)
4. \(\frac{1}{\sqrt{3}}\left(\frac{\tau}{B i}\right)\)

Answer: 2. \(2\left(\frac{\tau}{\sqrt{3} B i}\right)^{1 / 2}\)

The area of the triangle is

⇒ \(A=\frac{1}{2}(l)\left(\frac{\sqrt{3}}{2} l\right)=\frac{\sqrt{3}}{4} l^2\)

Magnetic moment of the current loop,

⇒ \(\vec{m}=i A=\frac{\sqrt{3}}{4} l^2 i\) is directed along the normal to its plane.

Thus, angle between \(\vec{B}\) and \(\vec{m}\) is 90° and the torque is

⇒ \(\tau=m B \sin 90^{\circ}=\frac{\sqrt{3}}{4} i l^2 B\)

∴ Side, l = \(l=\sqrt{\frac{4 \tau}{\sqrt{3} i B}}=2\left(\frac{\tau}{\sqrt{3 i B}}\right)^{1 / 2}\)

Question 3. A bar magnet having a magnetic moment of 2 x 10⁴ J T^-1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 x 10^-4 T exists in the space. The work done in taking the magnet slowly from the direction of the field to a direction 60° from the field is

1. 12 J
2. 6 J
3. 2J
4. 6.6 J

Answer: 2. 6 J

Given that magnetic moment = m = 2 x l0⁴ J T^-1, magnetic field = B = 6 x l0^-4 T^-1, deflection = θ = 60°.

Hence, work done is

∴ \(W=m B(1-\cos \theta)=\left(2 \times 10^4 \mathrm{~J} \mathrm{~T}^{-1}\right)\left(6 \times 10^{-4} \mathrm{~T}\right)\left(\frac{1}{2}\right)=6 \mathrm{~J}\)

Question 4. A short bar magnet of magnetic moment 0.4 J T^-1 is placed in a uniform magnetic field of 0.16 T. The magnetic is in stable equilibrium when the potential energy is

1. 0.064 J
2. -0.064 J
3. -0.082 J
4. Zero

Answer: 2. -0.064 J

The potential energy of a magnetic dipole in a magnetic field is

⇒ \(U=-\vec{m} \cdot \vec{B}=-m B \cos \theta\)

For stable equilibrium, PE must have a minimum obtained at θ = 0°.

PE in stable equilibrium is

∴ \(U=-m B \cos 0^{\circ}=-\left(0.4 \mathrm{~J} \mathrm{~T}^{-1}\right)(0.16 \mathrm{~T})=-0.064 \mathrm{~J}\)

Question 5. A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes angular SHM with a time period of 2 s in the earth’s horizontal magnetic field1 of 24 μ T. When a horizontal field of 18 μ T is produced opposite the earth’s field by placing a current-carrying wire, the new time period of the magnet will be

1. 1s
2. 2s
3. 3 s
4. 4 s

Answer: 4. 4 s

Time period = \(T=2 \pi \sqrt{\frac{I}{m B}}\)

Since I and m are constant,

⇒ \(T \propto \frac{1}{\sqrt{B}}\)

⇒ \(\frac{T_1}{T_2}=\sqrt{\frac{B_2}{B_1}}=\sqrt{\frac{B_1-B}{B_1}}=\sqrt{\frac{24 \mu \mathrm{T}-18 \mu \mathrm{T}}{24 \mu \mathrm{T}}}=\frac{1}{2}\).

But T₁ = 2 s, hence T₂ = 2T₁ = 4 s.

Question 6. A charged particle (charge q) is moving in a circle of radius R with a uniform speed v. The associated magnetic moment fi is given by

1. qvR2
2. \(\frac{q v R^2}{2}\)
3. \(\frac{q v R}{2}\)
4. qvR

Answer: 3. \(\frac{q v R}{2}\)

Current = I = \(\frac{q}{T}=\frac{q v}{2 \pi R}\)

The associated magnetic moment is

∴ \(m=I A=\frac{q v}{2 \pi R} \cdot \pi R^2=\frac{q v R}{2}\).

class 12th physics magnetism and matter

Question 7. If the file magnetic dipole moment of an atom of diamagnetic material, paramagnetic material, and ferromagnetic material is denoted by μ_p, μ_d,  and μ_f respectively then.

1. \(\mu_{\mathrm{d}}=0 \text { and } \mu_{\mathrm{p}} \neq 0\)
2. \(\mu_d \neq 0 \text { and } \mu_p=0\)
3. \(\mu_{\mathrm{p}}=0 \text { and } \mu_{\mathrm{f}} \neq 0\)
4. \(\mu_{\mathrm{d}} \neq 0 \text { and } \mu_{\mathrm{f}} \neq 0\)

Answer: 1. \(\mu_{\mathrm{d}}=0 \text { and } \mu_{\mathrm{p}} \neq 0\)

Diamagnetic substances do not have magnetic dipole moments \(\left(\mu_{\mathrm{d}}=0\right)\) and have negative susceptibility.

∴ However, paramagnetic magnets have positive magnetic moments. Thus, \(\mu_d=0, \mu_p \neq 0\)

Question 8. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in an equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is

1. \(\frac{w}{\sqrt{3}}\)
2. \(\sqrt{3} W\)
3. \(\frac{\sqrt{3} W}{2}\)
4. \(\frac{2 W}{\sqrt{3}}\)

Answer: 2. \(\sqrt{3} W\)

Work done to rotate the magnet by 60° is

⇒ \(W=m B(1-\cos \theta)=m B\left(1-\cos 60^{\circ}\right)=\frac{m B}{2}\)

mB = 2W.

The torque required to maintain the deflected position is

∴ \(\tau=m B \sin 60^{\circ}=2 W \cdot \frac{\sqrt{3}}{2}=\sqrt{3} W\).

Question 9. A bar magnet of magnetic moment m is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet, the length of the magnet will be

1. \(\frac{m B}{F}\)
2. \(\frac{B F}{m}\)
3. \(\frac{m F}{B}\)
4. \(\frac{F}{m B}\)

Answer: 1. \(\frac{m B}{F}\)

The torque on the magnet at 0 = 90° is τ = mBsin 90° = mB.

But, τ = F.l.

∴ Hence, length of the magnet = l = \(\frac{m B}{F}\)

Question 10. A magnetic needle suspended parallel to a magnetic field requires √3 f of work to turn it through 60. The torque needed to maintain it in this position will be

1. 23 J
2. 3J
3. 3 J
4. \(\frac{3}{2} \mathrm{~J}\)

Answer: 2. 3J

⇒ \(W=m B\left(1-\cos 60^{\circ}\right)=\frac{m B}{2}\)

The required torque to maintain the deflected position is

∴ \(\tau=m B \sin 60^{\circ}=(2 W) \frac{\sqrt{3}}{2}=\sqrt{3} W=\sqrt{3}(\sqrt{3} \mathrm{~J})=3 \mathrm{~J}\).

Question 11. There are four lightweight rod samples A, B, C, and D respectively. They are separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted.

• A is feebly repelled.
• B is feebly attracted.
• C is strongly attracted.
• D remains unaffected.

Which one of the following is true?

1. B is of a paramagnetic material.
2. C is of a diamagnetic material.
3. D is of a ferromagnetic material.
4. A is of a nonmagnetic material.

class 12th physics magnetism and matter

Answer: 1. B is of a paramagnetic material.

With a bar magnet,

A ferromagnetic material is strongly attracted,

A paramagnetic material is feebly attracted,

A diamagnetic material is feebly repelled and

A nonmagnetic material remains unaffected.

Hence, A is diamagnetic, B is paramagnetic, C is ferromagnetic and D is nonmagnetic.

Question 12. The work done in turning a magnet of magnetic moment m by an angle of 90° from the meridian is n times the corresponding work done to turn it through an angle of 60°. The value of n is given by

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. 2
4. 1

Answer: 3. 2

\(W_1=m B\left(1-\cos 90^{\circ}\right)=m B . \quad W_2=m B\left(1-\cos 60^{\circ}\right)=\frac{m B}{2}\).

Given, \(W_1=n W_2, \text { hence } m B=n \frac{m B}{2} \Rightarrow n=2\).

Question 13. To protect sensitive equipment from the external magnetic field, it should be

1. Surrounded by fine copper sheet
2. Placed inside an iron can
3. Wrapped with insulation around it while passing current through it
4. Placed inside an aluminum can

Answer: 2. Placed inside an iron can

When sensitive equipment is placed inside a soft iron can, it will be shielded from the influence of an external magnetic field.

Question 14. A bar magnet of magnetic moment \(\vec{m}\) is placed in a magnetic field of B. The torque exerted on it is

1. \(\vec{m}+\vec{B}\)
2. \(\vec{m} \cdot \vec{B}\)
3. \(-\vec{m} \cdot \vec{B}\)
4. \(-\vec{B} \times \vec{m}\)

Answer: 4. \(-\vec{B} \times \vec{m}\)

Torque on a magnet = \(\vec{\tau}=\vec{m} \times \vec{B}=-\vec{B} \times \vec{m}\).

Question 15. Two bar magnets having the same geometry with magnetic moments m and 2m are first placed in such a way that their like poles are together and their time period of oscillation is T₁.  Next, when the polarity of one of the magnets is reversed, their time period of oscillation becomes T₂. Thus,

1. T₁ < T₂
2. T₁ = T₂
3. T₁ > T₂
4. T₂ = ∞

Answer: 1. T₁ < T₂

Magnetic dipole moments are vectors.

When like poles are together, \(\vec{m}=\vec{m}_1+\vec{m}_2\) and when unlike poles are together, \(\vec{m}=\vec{m}_1-\vec{m}_2\).

Corresponding time periods are

⇒ \(T_1=2 \pi \sqrt{\frac{I}{\left(m_1+m_2\right) B}}, T_2=2 \pi \sqrt{\frac{I_1}{\left(m_1-m_2\right)}}\)

∴ T1 < T2.

Question 16. The following figures show the arrangement of two identical bar magnets in different configurations. Which arrangement has the highest net magnetic dipole moment?

1. 1
2. 2
3. 3
4. 4

Answer: 3. 3

Magnetic dipole moment is a vector quantity for which the resultant magnetic moment is

⇒ \(m=\sqrt{m_1^2+m_2^2+2 m_1 m_2 \cos \theta}\)

For

1. θ = 90°,
2. θ =180°,
3. θ = 30° and
4. θ = 60°.

Thus, m is the maximum for (3).

class 12th physics magnetism and matter

Question 17. A tape recorder records sound in the form of

1. Magnetic energy
2. Electrical energy
3. Variable resistance on the tape
4. Magnetic field on the tape

Answer: 4. Magnetic field on the tape

A tape recorder records sound in the form of a magnetic field on the tape. As the tape rubs against the recording head, it applies a magnetic field that is proportional to the input signal. The signal orients the magnetic particles in a specific format which acts as indicators of the signal stored.

Question 18. Domain formation is a necessary feature of

1. Ferromagnetism
2. Paramagnetism
3. Diamagnetism
4. All of these

Answer: 1. Ferromagnetism

The magnetic domain is a region within a ferromagnetic material in which the magnetization is in a uniform direction, so domain formation is an essential feature of ferromagnetism.

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Question 19. Field lines due to Earth’s horizontal magnetic field are

1. Concentric circles
2. Curved lines
3. Parallel and straight lines
4. Elliptical

Answer: 3. Parallel and straight lines

Earth’s magnetic field varies in magnitude and direction from place to place. The horizontal components are parallel and straight.

Question 20. Which of the following is true regarding diamagnetic substances (the symbols have their usual meaning)?

1. \(\mu_{\mathrm{r}}>1, \chi_m>1\)
2. \(\mu_{\mathrm{r}}>1, \chi_{\mathrm{m}}<1\)
3. \(\mu_{\mathrm{r}}<1, \chi_{\mathrm{m}}<0\)
4. \(\mu_r<1 ; \chi_m>1\)

Answer: 3. \(\mu_{\mathrm{r}}<1, \chi_{\mathrm{m}}<0\)

For a diamagnetic substance, susceptibility is negative \(\left(\chi_m<0\right)\) and relative permeability is less than unity \(\left(\mu_{\mathrm{r}}<1\right)\). This corresponds to option (3).

Question 21. A bar magnet made of steel has a magnetic moment of 2.5 A m² and a mass of 6.6 g. The intensity of magnetization of the magnet is (take the density of steel = 7.9 x 10³ kg m^-3 )

1. 1.0 x 10⁶ A m^-1
2. 2.0 x 10^-6 A m^-1
3. 3.0 x 10⁶ A m^-1
4. 3.0 x 10^-2 A m^-1

Answer: 3. 3.0 x 10⁶ A m^-1

Given, magnetic moment = m = 15 A m².

The volume of the sample is

⇒ \(V=\frac{\text { mass }}{\text { density }}=\frac{6.6 \times 10^{-3} \mathrm{~kg}}{7.9 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}=\frac{6.6}{7.9} \times 10^{-6} \mathrm{~m}^3\)

The intensity of magnetization is

∴ \(I=\frac{m}{V}=\frac{2.5 \mathrm{~A} \mathrm{~m}^2}{\left(\frac{66}{79} \times 10^{-6} \mathrm{~m}^3\right)}=\frac{2.5 \times 79}{66} \times 10^6 \mathrm{~A} \mathrm{~m}^{-1}=3 \times 10^6 \mathrm{~A} \mathrm{~m}^{-1}\).

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Question 22. A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb m^-2. The coil carries a steady current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be

1. 0.24 N m
2. 0.12 N m
3. 0.15 N m
4. 0.20 N m

Answer: 4. 0.20 N m

Area of the rectangular coil = A = (0.12 m)(0.1 m) =12 x10^-3 m²,

N = 50 turns, magnetic field = B = 0.2T, current = I = 2 A,

The angle between area vector and field = θ = 60°.

The required torque to maintain a stable equilibrium is

T = mBsin θ = (IAN)Bsin 60°

∴ \((2 \mathrm{~A})\left(12 \times 10^{-3} \mathrm{~m}^2\right)(50)(0.2 \mathrm{~T}) \frac{\sqrt{3}}{2}=0.20 \mathrm{~N} \mathrm{~m}\).

Question 23. A current loop in a magnetic field

1. Can be in equilibrium in two orientations but the equilibrium states are unstable
2. Can be in equilibrium in two orientations, one stable while the other unstable
3. Experiences a torque whether the field is uniform or nonuniform in all orientations
4. Can be in equilibrium in one orientation

Answer: 2. Can be in equilibrium in two orientations, one stable while the other unstable.

A current loop of magnetic moment \(\vec{m}\) can stay in equilibrium at two positions, one with θ = 0° and PE = -mB (= minimum) when equilibrium is stable, and other with θ =180° and PE = +mB(= maximum) when the equilibrium is unstable

Question 24. A bar magnet of length l and magnetic dipole moment m is bent in the form of an arc as shown in the figure. The new magnetic dipole moment will be

1. \(\frac{2 m}{\pi}\)
2. \(\frac{m}{2}\)
3. m
4. \(\frac{3 m}{\pi}\)

Answer: 4. \(\frac{3 m}{\pi}\)

The magnetic moment of the bar magnet is m = p_m l, where p_m = magnetic pole strength. When bent into the form of an arc,

⇒ \(l=\frac{\pi}{3} r \Rightarrow r=\frac{3 l}{\pi}\)

Hence, the magnetic moment is

∴ \(m^{\prime}=p_{\mathrm{m}} r=\frac{p_{\mathrm{m}} 3 l}{\pi}=\frac{3 m}{\pi}\).

Question 25. If θ₁ and θ₂ are the apparent angles of dip observed in two vertical planes mutually perpendicular to each other then the true angle of dip (θ) is given by

1. \(\tan ^2 \theta=\tan ^2 \theta_1+\tan ^2 \theta_2\)
2. \(\cot ^2 \theta=\cot ^2 \theta_1-\cot ^2 \theta_2\)
3. \(\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2\)
4. \(\tan ^2 \theta=\tan ^2 \theta_1-\tan ^2 \theta_2\)

Answer: 3. \(\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2\)

True dip θ at a place is given by

⇒ \(\tan \theta=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}}}\)

\(\cot ^2 \theta=\left(\frac{B_{\mathrm{H}}}{B_{\mathrm{V}}}\right)^2\) →(1)

In plane 1, inclined ata with the magnetic meridian,

⇒ \(\tan \theta_1=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}_1}}=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}} \cos \alpha}\)

⇒ \(\cot ^2 \theta_1=\left(\frac{B_{\mathrm{H}}}{B_{\mathrm{V}}}\right)^2 \cos ^2 \alpha\) →(2)

Similarly, in plane 2,

⇒ \(\cot ^2 \theta_2=\left(\frac{B_{\mathrm{H}}}{B_{\mathrm{V}}}\right)^2 \sin ^2 \alpha\) →(3)

Adding (2) and (3),

∴ \(\cot ^2 \theta_1+\cot ^2 \theta_2=\left(\frac{B_{\mathrm{H}}}{B_{\mathrm{V}}}\right)^2=\cot ^2 \theta\)

Question 26. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It will

1. Become rigid showing no movement
2. Stay in any position
3. Stay in the north-south direction only
4. Stay in the east-west direction only

Answer: 2. Stay in any position

At the geomagnetic pole, the magnetic field lines are vertical. So, the component of the magnetic field in the horizontal direction will be zero. Hence, the compass needle constrained to move in a horizontal plane will stay in any position.

class 12th physics magnetism and matter

Question 27. Two identical bar magnets are fixed with their centers at a distance d apart. A stationary charge Q is placed at P in the gap between the two magnets at a distance D from the center O as shown in the figure. The force on the charge Q is

1. Zero
2. Directed along OP
3. Directed along PO
4. Directed perpendicular to the plane of the paper

Answer: 1. Zero

Force on a charge q in a magnetic field B is \(\vec{F}=q(\vec{v} \times \vec{B})\).

Hence, the force on the stationary charge Q at P will be zero since ν = 0.

Question 28. Curie temperature is the temperature above which a

1. Paramagnetic material becomes ferromagnetic material
2. Ferromagnetic material becomes diamagnetic material
3. Ferromagnetic material becomes paramagnetic material
4. Paramagnetic material becomes diamagnetic material

Answer: 3. Ferromagnetic material becomes paramagnetic material

Curie temperature is the temperature above which a ferromagnetic substance loses its ferromagnetism and becomes paramagnetic.

Question 29. Nickel shows ferromagnetic properties at room temperature. If the temperature is increased beyond Curie’s temperature then it will show

1. Anti-ferromagnetism
2. No magnetic property
3. Diamagnetism
4. Paramagnetism

Answer: 4. Paramagnetism

Nickel (a ferromagnetic substance) when heated above the Curie temperature, becomes paramagnetic.

Question 30. In which of the following, the magnetic susceptibility does not depend on temperature?

1. Diamagnetism
2. Ferromagnetism
3. Paramagnetism
4. None of these

Answer: 1. Diamagnetism

Magnetic susceptibility of a diamagnetic substance is small and negative \((x<0)\) which does not depend on temperature.

Question 31. The magnetic moment of a diamagnetic atom is

1. Much greater than one
2. One
3. Between zero and one
4. Equal to zero

Answer: 4. Equal to zero

In diamagnetic materials all the electrons are paired so there is no permanent magnetic moment per atom.

Question 32. The magnetic susceptibility is negative for

1. Ferromagnetic materials only
2. Paramagnetic and ferromagnetic materials
3. Diamagnetic materials only
4. Paramagnetic materials only

Answer: 3. Diamagnetic materials only

Magnetic susceptibility for diamagnetic materials is negative.

Question 33. Electromagnets are made of soft iron because soft iron has

1. Low retentivity and high coercivity
2. High retentivity and high coercivity
3. Low retentivity and low coercivity
4. High retentivity and low coercivity

Answer: 3. Low retentivity and low coercivity

An electromagnet retains its magnetism as long as current flows through its windings and becomes an ordinary piece of iron in the absence of current.

Hence, electromagnets are made of soft iron which has low retentivity and coercivity.

Question 34. If a diamagnetic substance is brought near the north or south pole of a bar magnet, it is

1. Repelled by the North Pole and attracted by the South Pole
2. Attracted by the North Pole and repelled by the South Pole
3. Attracted by both poles
4. Repelled by both poles

Answer: 4. Repelled by both the poles

A diamagnetic substance moves from the stronger part of the magnetic field to the weaker part. Hence, it will be repelled by both the magnetic poles of a magnet.

Question 35. According toCurie’slaw, the magnetic susceptibility χ of a magnetic substance is proportional to absolute temperature T as

1. \(\frac{1}{T}\)
2. T
3. \(\frac{1}{T^2}\)
4. \(T^2\)

Answer: 1. \(\frac{1}{T}\)

According to Curie’s law, susceptibility \((\chi)\) is inversely proportional to temperature (T), thus \(\chi \propto \frac{1}{T}\).

Question 36. The angle of dip at a certain place where the horizontal and vertical components of the earth’s magnetic field are equal is

1. 30°
2. 75°
3. 60°
4. 45°

Answer: 4. 45°

The angle of dip (δ) at a place is given by tan \(\delta=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}}}\), where By and Bn are BH the components of a magnetic field along the vertical and horizontal directions. Given, B_v = B_H; so tan δ =1 ⇒ δ = 45°.

Question 37. The angles of dip at the poles and the equator respectively are

1. 30° and 60°
2. 0° and 90°
3. 45° and 90°
4. 90° and 0°

Answer: 4. 90° and 0°

A freely suspended needle always aligns its axis parallel to the magnetic field. At geomagnetic poles, field lines are vertical, so dip δ = 90°. At the equator, field lines are horizontal, so dip δ = 0°

Question 38. The temperature of transition from ferromagnetic property to paramagnetic property is called

1. Transition temperature
2. Critical temperature
3. Curie temperature
4. Triple temperature

Answer: 3. Curie temperature

Curie temperature is the temperature at which a ferromagnetic material changes to a paramagnetic material.

Question 39. The relation between magnetic moment m and angular velocity is

1. \(m \propto \omega\)
2. \(m \propto \sqrt{\omega}\)
3. \(m \propto \omega^2\)
4. None of these.

Answer: 1. \(m \propto \omega\)

The magnetic moment of a current loop is m = IA.

For a charge revolving in a circular orbit,

⇒ \(m=I A=\frac{q}{T} \pi r^2=q\left(\frac{v}{2 \pi r}\right) \pi r^2\)

⇒ \(q \frac{\omega r^2}{2}=\left(\frac{1}{2} q r^2\right) \omega\).

Thus, m ∝ CD.

Question 40. Due to the earth’s magnetic field, charged cosmic ray particles

1. Can never reach the poles
2. Can never reach the equator
3. Require less kinetic energy to reach the equator
4. Requires greater kinetic energy to reach the equator than the poles

Answer: 4. Require greater kinetic energy to reach the equator than the poles

The earth’s magnetic field is along the vertical at the poles and along
the horizontal at the equator. At the poles, the magnetic force on charged particles \((\vec{F}=q \vec{v} \times \vec{B})\) is zero since the angle between \(\vec{V}\) and \(\vec{B}\) is 0. But at the equator, v and B are mutually perpendicular, i.e., θ = 90° and the deflecting force is maximum. Thus, only particles with greater KE can reach the equator.

Question 41. At point A on the earth’s surface the angle of dip δ = +25°. At point B on the earth’s surface the angle of dip δ = -25°. We can conclude that

1. A and B are both located in the northern hemisphere
2. A is located in the southern hemisphere and B is located in the northern hemisphere
3. A is located in the northern hemisphere and B is located in the southern hemisphere
4. A and B are both located in the southern hemisphere.

Answer: 3. A is located in the northern hemisphere and B is located in the southern hemisphere

The dip is called positive if the north pole of the magnetic needle faces downward and negative if it points upward. The dip is zero at the equator and increases positively in the. northern hemisphere and negatively in the southern hemisphere as one approaches the magnetic pole where it is 90°. Hence, the dip is positive in the northern hemisphere (place A) and negative in the southern hemisphere (place B).

class 12th physics magnetism and matter

Question 42. A perfectly diamagnetic sphere has a small concentric spherical cavity at its center which is filled with a paramagnetic substance. The whole system is placed in a uniform magnetic field \(\vec{B}\). The field inside the paramagnetic substance is

1. \(\vec{B}\)
2. Much stronger than \(|\vec{B}|\) I but opposite to \(\vec{B}\)
3. Much larger than \(\vec{B}\) and parallel to \(\vec{B}\)
4. Zero

Answer: 4. Zero

Diamagnetic substances do not allow magnetic field lines to pass through them. Hence, the field inside is zero.

Question 43. An iron rod of susceptibility 599 is subjected to a magnetizing field of 1200 A m^-1. The permeability of the material is \(\left(\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}\right)\)

1. \(8.0 \times 10^{-5} \mathrm{~T} \mathrm{~mA}^{-1}\)
2. \(2.4 \pi \times 10^{-5} \mathrm{~T} \mathrm{~mA}^{-1}\)
3. \(2.4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}\)
4. \(2.4 \pi \times 10^{-4} \mathrm{~T} \mathrm{~mA}^{-1}\)

Answer: 4. \(2.4 \pi \times 10^{-4} \mathrm{~T} \mathrm{~mA}^{-1}\)

Relative permeability = \(\left(\frac{\mu}{\mu_0}\right)=\mu_{\mathrm{r}}=1+\chi, \text { where } \chi=\text { susceptibility }\)

⇒ \(\mu=\mu_0(1+\chi)\)

⇒  (4π x 10^-7 T m A^-1)(1 + 599) .

⇒  (600)(4π)(10^-7 T m A^-1)

∴ 2.4π x lO^-4T m A^-1.

Question 44. There are two magnets P and T. P is used as a permanent magnet while T is used in transformers. Then,

1. P has high retentivity and low coercivity
2. P has low retentivity and high coercivity
3. T has low coercivity and low retentivity
4. T has high coercivity and high retentivity

Answer: 3. T has low coercivity and low retentivity

Steel has high coercivity and high retentivity due to which it is used in the making of permanent magnets.

Soft iron has low retentivity and low coercivity due to which it is used in the making of laminated cores in transformers and electromagnets.

Question 45. The controlling torque on a bar magnet making an angle of 30° with the direction of the uniform magnetic field of strength 0.06 T is 0.018 N m. Find the work done by an external force in deflecting it from the configuration of minimum potential energy to that of maximum potential energy.

1. 0.036 J
2. 0.018 J
3. 0.072 J
4. 0.18 J

Answer: 3. 0.072 J

Torque = t = mBsin θ

⇒ \(m=\frac{\tau}{B \sin \theta}=\frac{0.018 \mathrm{Nm}}{0.06 \mathrm{~T} \times \sin 30^{\circ}}\)

= \(\frac{0.018}{0.06 \times 0.5}=0.6 \mathrm{~A} \mathrm{~m}\)

Potential energy is

⇒ \(U=-\vec{m} \cdot \vec{B}=-m B \cos \theta\)

⇒ \(U_{\min }=-m B \cos 0^{\circ}=-m B\)

⇒ \(U_{\max }=-m B \cos \pi=+m B\)

work done by the external agent is

∴ W = ΔU = 2mB = 2(0.6 A m)(0.06 T) = 0.072 J.

Question 46. An iron rod of volume 10^-3 m³ and relative permeability 1000 is placed as a core in a solenoid with 10 turns cm^-1. If a steady current of 0.5 A is passed through the solenoid then the magnetic moment of the solenoid will be

1. 500 A m²
2. 500 x 10² A m²
3. 0.5 x 10² A m²
4. 5000 A m²

Answer: 1. 500 A m²

B = μH =μ_rμ₀H = μ₀(H+I),

where I = magnetization = \(\frac{\text { magnetic moment }(m)}{\text { volume }(V)}\)

μ_rH = H+I ⇒ I = (μ_r-1)H = (μ_r– l)nI

⇒ \(\frac{m}{V}=\left(\mu_r-1\right) n I\)

⇒ m = (1000-1)(10 x100 m^-1)(0.5 A)(10^-3 m³)

∴ (999)(10³ x 0.5)(10^-3) A m² ≈ 500 A m².

Question 47. A paramagnetic sample shows a net magnetization of 6 A m^-1 when it is placed in an external magnetic field of 0.4 T at a temperature of 4 K. When placed in an external field of 0.3 T at a temperature of 24 K, the magnetization will be

1. 0.75 A m^-1
2. 2.25 A m^-1
3. 1.75 A m^-1
4. 4 A m^-1

Answer: 1. 0.75 A m^-1

According to Curie’s law,

susceptibility \(\chi \propto \frac{1}{T}\)

⇒ \(\frac{I}{H} \propto \frac{1}{T}, \text { hence } \frac{I T}{H}=\text { constant }\)

∴ \(\frac{I_1 T_1}{H_1}=\frac{I_2 T_2}{H_2} \Rightarrow \frac{\left(6 \mathrm{~A} \mathrm{~m}^{-1}\right)(4 \mathrm{~K})}{(0.4 \mathrm{~T})}=\frac{I_2(24 \mathrm{~K})}{(0.3 \mathrm{~T})} \Rightarrow I_2=0.75 \mathrm{~A} \mathrm{~m}^{-1}\)