Q: If the magnetic susceptibilities of diamagnetic, paramagnetic, and ferromagnetic substances are denoted by χd, χp, and χf respectively, which ordering is correct?

χd < χp < χf. Diamagnetic materials have a negative susceptibility, indicating they are repelled by magnetic fields, while paramagnetic materials have a small positive susceptibility, meaning they are weakly attracted to magnetic fields. Ferromagnetic materials exhibit a much larger positive susceptibility, as they can be strongly attracted and retain magnetization, thus establishing the order χd < χp < χf.

Q: At a certain location, the magnetic dip is 37° and the vertical component of Earth’s magnetic field is 6 × 10⁻⁵ T. Taking tan 37° = 3/4, what is the magnitude of Earth’s magnetic field there?

10⁻⁴ T. The magnitude of Earth's magnetic field can be calculated using the relationship between the vertical component and the angle of dip. Given that the vertical component is 6 × 10⁻⁵ T and using the tangent of the dip angle, we find that the total magnetic field is approximately 10⁻⁴ T.

Question 1

The SI unit Weber is used for measuring which of the following quantities?

Accepted Answer

magnetic flux. The weber (Wb = V*s = T*m^2) is the SI unit of magnetic flux.

Question 2

Two identical magnetic dipoles, each having magnetic moment 1.0 A·m², are kept 2 m apart with their axes mutually perpendicular. The magnetic field at the midpoint between them is

Accepted Answer

√5 × 10⁻⁷ T. At the midpoint (1 m from each), one dipole gives an axial field 2*(1e-7)*m = 2e-7 T and the other an equatorial field (1e-7)*m = 1e-7 T. As the axes are perpendicular these add in quadrature: sqrt(4+1)*1e-7 = sqrt(5)*1e-7 T.

Question 3

A material has relative permittivity εr and relative permeability μr. For a diamagnetic substance, which pair of values is physically permissible?

Accepted Answer

εr = 1.5, μr = 0.5. The correct option indicates that the relative permittivity can be greater than 1 while the relative permeability is less than 1, which is consistent with the properties of diamagnetic materials that exhibit weak repulsion to magnetic fields and can have varying electric properties.

Question 4

A bar magnet freely suspended in the horizontal component of Earth’s magnetic field H has a time period of 4 s. If a second magnet is placed nearby and the time period becomes 2 s, then the magnetic field produced by the second magnet is:

Accepted Answer

3H. Since T is proportional to 1/sqrt(field), (T1/T2)^2 = 4 gives a net field of 4H. The added field of the second magnet is 4H - H = 3H.

Question 5

Which magnetic properties should the core material possess to be suitable for an electromagnet?

Accepted Answer

Low retentivity and low coercivity. An electromagnet must lose magnetism when current stops (low retentivity) and be easily demagnetized (low coercivity). Soft iron has both, so the answer is low retentivity and low coercivity.

Question 6

A bar magnet has a length much greater than its width and thickness. In a vibration magnetometer, its oscillation period is 2 s. The magnet is then divided lengthwise into three equal pieces, and the three pieces are stacked together so that like poles are in contact. What is the oscillati

Accepted Answer

2 s. The oscillation period of a magnet in a vibration magnetometer depends on its moment of inertia and the restoring torque, which are unchanged when the magnet is divided and reassembled with like poles together. Since the effective magnetic moment remains the same, the oscillation period remains at 2 s.

Question 7

A magnetic dipole is placed in two magnetic fields whose directions differ by 60°. If one field has magnitude 1.2 × 10⁻² T and the dipole settles in stable equilibrium making an angle of 15° with that field, what is the magnitude of the second field?

Accepted Answer

4.4 × 10⁻³ tesla. In stable equilibrium the torques balance: B1*sin(theta1) = B2*sin(theta2) with theta1 = 15 deg and theta2 = 60 - 15 = 45 deg. So B2 = 1.2e-2 * sin15/sin45 = 4.4e-3 T.

Question 8

A bar magnet of length 8 cm is aligned along the magnetic meridian with its north pole directed toward geographic north. On the equatorial line of the magnet, two neutral points are found, and the distance between them is 6 cm. If the horizontal component of Earth’s magnetic field is 3.2 ×

Accepted Answer

5 ab-amp × cm. Neutral points lie 3 cm from centre (6/2) on the equatorial line, half-length l = 4 cm. Equatorial field M/(r^2+l^2)^1.5 = B_H gives M*1/125 = 0.32 G, so M = 40 G*cm^3. With length 8 cm, pole strength m = M/8 = 5 ab-amp*cm.

Question 9

If the magnetic susceptibilities of diamagnetic, paramagnetic, and ferromagnetic substances are denoted by χd, χp, and χf respectively, which ordering is correct?

Accepted Answer

χd < χp < χf. Diamagnetic materials have a negative susceptibility, indicating they are repelled by magnetic fields, while paramagnetic materials have a small positive susceptibility, meaning they are weakly attracted to magnetic fields. Ferromagnetic materials exhibit a much larger positive susceptibility, as they can be strongly attracted and retain magnetization, thus establishing the order χd < χp < χf.

Question 10

At a certain location, the magnetic dip is 37° and the vertical component of Earth’s magnetic field is 6 × 10⁻⁵ T. Taking tan 37° = 3/4, what is the magnitude of Earth’s magnetic field there?

Accepted Answer

10⁻⁴ T. The magnitude of Earth's magnetic field can be calculated using the relationship between the vertical component and the angle of dip. Given that the vertical component is 6 × 10⁻⁵ T and using the tangent of the dip angle, we find that the total magnetic field is approximately 10⁻⁴ T.

Question 11

Needles N1, N2 and N3 are respectively made of ferromagnetic, paramagnetic and diamagnetic materials. If a magnet is brought near them, it will

Accepted Answer

pull N1 strongly, pull N2 weakly, and push away N3 weakly. A ferromagnetic needle is attracted strongly, a paramagnetic one is attracted weakly, and a diamagnetic one is repelled weakly. This matches: pull N1 strongly, pull N2 weakly, push away N3 weakly.

Question 12

A vibration magnetometer has two identical bar magnets arranged one above the other so that they are mutually perpendicular and each bisects the other. When this system oscillates in a horizontal magnetic field, its time period is 2^(3/4) s. If one magnet is taken away and the remaining ma

Accepted Answer

2^(1/2). Two perpendicular identical magnets: net moment = sqrt2*M, inertia = 2I, so T_two = 2*pi*sqrt(2I/(sqrt2*M*B)) = 2^(1/4)*T_one. Thus T_one = 2^(3/4)/2^(1/4) = 2^(1/2) s.

Question 13

At the Earth’s equator, the magnetic field is about 4 × 10⁻⁵ T. If the Earth’s radius is 6.4 × 10⁶ m, then its magnetic dipole moment is closest to which order of magnitude?

Accepted Answer

10²³ A m². Equatorial field B = (mu0/4pi) M/r^3, so M = B r^3/(1e-7) = 4e-5 * (6.4e6)^3 / 1e-7 ~ 1.05e23 A m^2, i.e. order 10^23.

Question 14

Which statement about relative magnetic permeability (μr) is incorrect?

Accepted Answer

In paramagnetic substances, μr is less than 1.. The statement is incorrect because in paramagnetic substances, the relative magnetic permeability (μr) is actually slightly greater than 1, indicating that they have a weak positive response to an external magnetic field.

Question 15

A bar magnet with moment of inertia 9 × 10⁻⁵ kg m² is suspended in a vibration magnetometer and performs oscillations in a uniform magnetic field of 16π² × 10⁻⁵ T. If it completes 20 oscillations in 15 s, what is the magnetic moment of the bar magnet?

Accepted Answer

4 Am². T = 15/20 = 0.75 s. From T = 2*pi*sqrt(I/(M*B)): M = 4*pi^2*I/(T^2*B) = 4*pi^2*9e-5/(0.5625*16*pi^2*1e-5) = 4 A*m^2.

Question 16

Two small bar magnets P and Q are placed far apart with their centres lying on the X-axis. The magnetic axis of P is along the X-axis, while that of Q is along the Y-axis. At point R, which is exactly midway between the two centres, let B denote the magnitude of the magnetic induction prod

Accepted Answer

√5B. The resultant magnetic induction at point R is found by vectorially adding the magnetic fields from both magnets. Since magnet P's field is along the X-axis and magnet Q's field is along the Y-axis, the resultant forms a right triangle, leading to the calculation of the resultant magnitude as √(B² + B²) = √5B.

Question 17

A dip circle is adjusted so that its needle is free to move in the magnetic meridian, and the angle of dip at that place is 40°. The instrument is then turned so that the plane of the needle makes an angle of 30° with the magnetic meridian. In this new position, the needle will dip by an a

Accepted Answer

greater than 40°. When the plane makes angle 30 deg with the meridian, tan(dip') = tan(40)/cos(30) > tan(40), so the apparent dip is greater than 40 degrees.

Question 18

A watch glass filled with a powdered material is kept between the poles of a magnet. A pronounced depression appears at the middle. The material in the watch glass is

Accepted Answer

carbon. Carbon, being a non-magnetic material, does not respond to the magnetic field, leading to the formation of a depression in the powdered material as the magnetic field lines concentrate in the surrounding areas.

Question 19

A magnetic needle initially aligned parallel to a magnetic field needs an amount of work equal to W to be rotated by 60°. The torque needed to hold the needle at this angular position is

Accepted Answer

√3W. Work to rotate from 0 to 60 deg: W = MB(cos0 - cos60) = MB(1 - 0.5) = MB/2, so MB = 2W. Holding torque at 60 deg = MB*sin60 = 2W*(sqrt(3)/2) = sqrt(3)*W.

Question 20

Earth’s magnetic field can be approximated by that of a dipole located at its center. If this dipole has a magnetic moment of about 8 × 10²² A m², then the magnetic field strength near the Earth’s equator (take Earth’s radius as 6.4 × 10⁶ m) is approximately

Accepted Answer

0.32 Gauss. B = 1e-7 * 8e22 / (6.4e6)^3 = 1e-7 * 8e22 / 2.62e20 = 3.05e-5 T = 0.305 Gauss, i.e. about 0.32 Gauss.

Question 21

Two small bar magnets, each 1 cm long, have magnetic moments of 1.20 A m² and 1.00 A m². They are kept on a level table side by side, parallel to one another, with their north poles directed toward the south. Their magnetic equators coincide, and the distance between their centres is 20.0

Accepted Answer

2.56 × 10⁻⁴ Wb/m². O lies on the equatorial line of both magnets at d=0.1 m. B = (mu0/4pi)M/d^3 gives 1.2e-4 and 1.0e-4 T, both directed north (opposite to the southward moments), adding to Earth's 0.36e-4 T: total ~ 2.56e-4 Wb/m^2.

Question 22

A small magnet has a coercivity of 3 × 10³ A m⁻¹. What current must flow through a solenoid of length 10 cm with 100 turns so that the magnetic field inside it just demagnetizes the magnet placed within the solenoid?

Accepted Answer

3 A. Turns per metre n = 100/0.10 = 1000 /m. To demagnetize, the solenoid field H must equal the coercivity: n*I = 3e3, so I = 3000/1000 = 3 A.

Question 23

A thin rectangular bar magnet is freely suspended and executes small oscillations with period T. The magnet is then cut into two identical parts, each having half of the original length. If one of these halves is suspended freely in the same magnetic field and allowed to oscillate, its per

Accepted Answer

1/2. Cutting into halves of half length: moment of inertia becomes I/8 and magnetic moment becomes M/2. T' = 2*pi*sqrt((I/8)/((M/2)*B)) = 2*pi*sqrt(I/(4*M*B)) = T/2, so T'/T = 1/2.

Question 24

Identify the statement that is not true regarding hysteresis.

Accepted Answer

The area enclosed by the hysteresis loop does not depend on the heat produced per unit volume of the material.. This statement is correct because the area of the hysteresis loop is directly related to the energy loss due to magnetic hysteresis, which translates to heat produced per unit volume; therefore, it does depend on this heat production.

Question 25

A slender rectangular bar magnet is hung so that it can oscillate freely in a uniform magnetic field, and its time period is T. The magnet is then cut into two identical parts, each having half the original length. If one of these halves is suspended freely in the same field and allowed to

Accepted Answer

1/2. When the bar magnet is cut in half, its length is reduced, which affects its moment of inertia. The time period of oscillation is proportional to the square root of the length of the magnet, leading to a new time period that is half of the original.

Question 26

A magnetic needle initially aligned parallel to a uniform magnetic field is rotated by 60°. If the work done in this process is W, then the torque required to keep the needle fixed at this angle is

Accepted Answer

√(3)W. The torque required to maintain the needle at an angle in a magnetic field is proportional to the sine of the angle between the needle and the field. When the needle is rotated by 60°, the torque becomes sin(60°) times the work done, which results in ( rac{ ext{√3}}{2})W, leading to the conclusion that the torque is ext{√3}W.

Question 27

Inside a bar magnet, the magnetic field lines are directed

Accepted Answer

from the south pole to the north pole of the magnet. Inside a bar magnet, the magnetic field lines flow from the south pole to the north pole, which is the opposite direction of the external field lines. This is a fundamental characteristic of magnetic fields, where the lines always complete a loop from north to south outside the magnet and south to north inside.

Question 28

The Curie temperature is defined as the temperature beyond which

Accepted Answer

a ferromagnetic substance changes into a paramagnetic one. The Curie temperature marks the point at which a ferromagnetic material loses its permanent magnetism and becomes paramagnetic, meaning it can no longer maintain a magnetic alignment without an external magnetic field.

Question 29

A bar magnet has a length much greater than its width and thickness. In a vibration magnetometer, it oscillates with a period of 2 s. The magnet is then divided lengthwise into three identical pieces, and these pieces are stacked together so that like poles are aligned. What will be the ne

Accepted Answer

2 s. Each lengthwise strip has the same length, mass m/3 and moment M/3. Stacked with like poles aligned, total moment = M, total mass = m, total moment of inertia = m L^2/12 = original I. Since T = 2 pi sqrt(I/(M B)) is unchanged, the period stays 2 s.

Question 30

Relative permittivity and permeability of a material εr and μr respectively. Which of the following values of these quantities are allowed for a diamagnetic material?

Accepted Answer

εr = 1.5, μr = 0.5. Diamagnetic materials have a relative permeability (μr) less than 1, indicating they are repelled by magnetic fields, while their relative permittivity (εr) can be greater than 1, reflecting their ability to store electric energy. Therefore, the combination of εr = 1.5 and μr = 0.5 is consistent with the properties of a diamagnetic material.

Question 31

Two short bar magnets of length 1 cm each have magnetic moments 1.20 A m² and 1.00 A m² respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance 20.0 cm. The

Accepted Answer

2.56 × 10⁻⁴ Wb/m². At the common equatorial mid-point each magnet gives B = (mu0/4pi)*m/r^3 with r=0.10 m: 1.2e-4 and 1.0e-4 Wb/m^2, both directed the same way (N->S), summing to 2.2e-4. Adding earth's horizontal component 0.36e-4 gives about 2.56e-4 Wb/m^2.

Question 32

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 10³ A m⁻¹. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is:

Accepted Answer

3 A. Demagnetizing field H = (N/L)*I, so I = H*L/N = (3e3 * 0.10)/100 = 3 A.

Question 33

A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is:

Accepted Answer

2600 A/m. H = (N/L) I = (100/0.2) x 5.2 = 500 x 5.2 = 2600 A/m.

Question 34

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 10³ A m⁻¹. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized inside the solenoid, is -

Accepted Answer

3 A. The correct option is right because the current required to demagnetize the small magnet can be calculated using the formula for the magnetic field inside a solenoid, which is proportional to the product of the current and the number of turns per unit length. Given the coercivity and the dimensions of the solenoid, a current of 3 A is necessary to achieve the required magnetic field to demagnetize the magnet.

Question 35

A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is:

Accepted Answer

2600 A/m. Coercivity equals the demagnetizing field H = n*I = (N/L)*I = (100/0.2)*5.2 = 500*5.2 = 2600 A/m.

Question 36

Magnetic materials used for making permanent magnets (P) and magnets in a transformer (T) have different properties. Which property best matches for the type of magnet required?
(1) T: Large retentivity, small coercivity
(2) P: Small retentivity, large coercivity
(3) T: Large retentivity,

Accepted Answer

P: Large retentivity, large coercivity. A permanent magnet (P) must retain magnetisation and resist demagnetisation, so it needs large retentivity and large coercivity. A transformer core (T) is a soft magnet needing low retentivity and low coercivity. Hence the correct match is 'P: Large retentivity, large coercivity'.

Question 37

A small bar magnet placed with its axis at 30° with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is-

Accepted Answer

7.2 × 10⁻² J. From torque, mB*sin30 = 0.018 so mB = 0.036. Work from stable (0 deg) to unstable (180 deg) is mB(cos0 - cos180) = 2mB = 0.072 J = 7.2 x 10^-2 J.

Question 38

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

Accepted Answer

0.75 A/m. For a paramagnet M is proportional to B/T. M2 = 6 * (0.3/0.4) * (4/24) = 6 * 0.75 * (1/6) = 0.75 A/m.

Question 39

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

Accepted Answer

5 × 10² A m². The magnetic moment of the rod is calculated using the formula M = nIA, where n is the number of turns per unit length, I is the current, and A is the cross-sectional area of the rod. Given the parameters, the calculations yield a magnetic moment of 5 × 10² A m², confirming option D as the correct answer.

Question 40

Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by ϕi. The magnetic flux through the area of the circular coil is given by ϕo.

Accepted Answer

(4) ϕi = -ϕo. All field lines passing up through the coil return down through the rest of the infinite plane, so the total flux through the plane is zero. Hence phi_i = -phi_o.

Question 41

A bar magnet of length 14 cm is placed in magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the center of the magnet. If B_H = 0.4 G, the magnetic moment of the magnet is: (1 G = 10⁻⁴ T)

Accepted Answer

2.880 J T⁻¹. With N pole toward geographic north the neutral points lie on the equatorial line. So 0.4e-4 = 1e-7 * M/(0.18^2+0.07^2)^{3/2}. (0.0373)^{3/2}=7.21e-3, giving M = 4e-5*7.21e-3/1e-7 = 2.88 J/T.

Question 42

Which of the following statements are correct?
(A) Electric monopoles do not exist whereas magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its ends and outside cannot be completely straight and confined.
(C) Magnetic field lines are completely confined within a toro

Accepted Answer

(B) and (C) only. (A) is false (electric monopoles exist, magnetic ones do not). (B) is true: solenoid end/outside lines cannot be perfectly straight and confined. (C) is true: an ideal toroid confines its field lines. (D) is false (lines inside a bar magnet are essentially parallel). So the correct choice is (B) and (C) only.

Question 43

At an angle of 30° to the magnetic meridian, the apparent dip is 45°. Find the true dip:

Accepted Answer

tan⁻¹ (√3/2). The true dip can be calculated using the relationship between the apparent dip and the angle to the magnetic meridian. Given that the apparent dip is 45° and the angle to the magnetic meridian is 30°, the true dip is found using the formula that relates these angles, leading to the result of tan⁻¹(√3/2).

Question 44

The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is 4π × 10⁻⁷ H/m. Absolute permeability of the material of the rod is:

Accepted Answer

2π × 10⁻⁴ H/m. The absolute permeability of a material can be calculated using the formula: μ = μ₀ (1 + χ), where μ₀ is the permeability of vacuum and χ is the magnetic susceptibility. Given that the susceptibility is 499, the absolute permeability becomes 4π × 10⁻⁷ H/m × (1 + 499), which simplifies to 2π × 10⁻⁴ H/m.

Question 45

Statement I: The ferromagnetic property depends on temperature. At high temperature, ferromagnet becomes paramagnet.
Statement II: At high temperature, the domain wall area of a ferromagnetic domain increases.
In the light of the above statements, choose the most appropriate answer from th

Accepted Answer

Statement I is true but Statement II is false. Statement I is correct because ferromagnetic materials lose their magnetic properties and become paramagnetic at high temperatures due to thermal agitation disrupting the alignment of magnetic domains. Statement II is false because at high temperatures, the domain wall area does not increase; instead, the thermal energy causes the domains to become less ordered, leading to a decrease in overall magnetization.

Question 46

A soft ferromagnetic material is placed in an external magnetic field. The magnetic domains

Accepted Answer

may increase or decrease in size and change its orientation.. When a soft ferromagnetic material is subjected to an external magnetic field, the magnetic domains can realign and adjust in size to enhance the material's magnetization, allowing for both increases and decreases in domain size and changes in orientation.

Question 47

A bar magnet having a magnetic moment of 2.0 × 10⁵ J T⁻¹, is placed along the direction of uniform magnetic field of magnitude B = 14 × 10⁻⁵ T. The work done in rotating the magnet slowly through 60° from the direction of field is:

Accepted Answer

14 J. The work done in rotating a magnetic moment in a magnetic field is given by the formula W = -mB(cosθ - cosθ0). In this case, the initial angle θ0 is 0° and the final angle θ is 60°, leading to a calculation that results in 14 J.

Question 48

A magnet hung at 45° with magnetic meridian makes an angle of 60° with the horizontal. The actual value of the angle of dip is

Accepted Answer

tan⁻¹(√(3/2)). For a vertical plane at angle alpha to the magnetic meridian, tan(apparent) = tan(true)/cos(alpha). So tan(true) = tan60 * cos45 = sqrt(3)*(1/sqrt2) = sqrt(3/2). True dip = tan^-1(sqrt(3/2)).

Question 49

The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m, is [Permeability of free space μ0 = 4π × 10⁻⁷ Wb/A-m]

Accepted Answer

4π × 10⁻⁵. The permeability of a paramagnetic material can be calculated using the formula μ = μ0(1 + χ), where χ is the susceptibility. Given that the susceptibility is 99, the calculation yields a permeability of approximately 4π × 10⁻⁵ Wb/A-m.

Question 50

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 × 10⁻⁵. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

Accepted Answer

1.2 × 10⁻⁵. The fractional increase in the magnetic field inside the solenoid is directly related to the magnetic susceptibility of the material. Since the susceptibility is given as 1.2 × 10⁻⁵, this value represents the ratio of the increase in magnetic field strength compared to the air medium, making option A the correct choice.

JEE Main Physics: Magnetism and Matter questions with solutions

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