Important Questions for CBSE Class 12 Physics Chapter 5 Magnetism and Matter

Magnetism And Matter Multiple Choice Questions And Answers

Important Questions for CBSE Class 12 Physics Chapter 5 Magnetism and Matter

Question 1. The vertical component of the earth’s magnetic field at a place is \(\frac{1}{\sqrt{3}}\) times the horizontal component. The angle trip of dip at that place is:

  1. 30°
  2. 45°
  3. 60°

Answer: 2. 30°

Question 2. Inside a bar magnet, the magnetic field lines

  1. Are not present
  2. Are parallel to the cross-sectional area
  3. Are in the direction from N-poIcs to S-pole
  4. Are in the direction from S-pole to N-pole

Answer: 4. Are in the direction from S-pole to N-pole

Question 3. A magnetic needle is kept nonparallel to the magnetic field in non-uniform magnetic field experiences.

  1. A force but not a torque
  2. Torque but not a force
  3. Both a force and a torque
  4. Neither a force nor a torque

Answer: 3. Both a force and a torque

Question 4. When a paramagnetic substance is brought near a north pole or a south pole of a bar magnet, it __________.

  1. Experience repulsion
  2. Docs do not experience attraction or repulsion
  3. Experience attraction
  4. Experience attraction or repulsion depending upon which pole is brought near to it.

Answer: 3. Experience attraction

Initially m’ = qm x l

Magnetism And Matter Bar magnet

after bending

m’ = qm x 2r

⇒ \(m^{\prime}=\frac{m}{l} \times 2 \times \frac{l}{\pi}\)  [ πr = l ]

∴ \(m^{\prime}=\frac{2 m}{\pi}\)

question 5. A straight steel wire of length l has magnetic moment m. If the wire is bent in the form of a semicircle, the new value of the magnetic dipole moment is.

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

Answer: 4. \(\frac{2 \mathrm{~m}}{\pi}\)

Magnetism And Matter A Straight Steel Wire

Initially

m = qm x l

⇒ \(\mathrm{m}^{\prime}=\mathrm{q}_{\mathrm{m}} \times \frac{l}{2}\)

(qm will not change)

⇒ \(\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2}\)

∴ \(\vec{P}\) remains constant.

Question 6. A bar magnet of length l, pole strength ‘P’, and magnetic moment ‘\(\vec{m}\)‘ is split into two equal pieces each of length l/2. The magnetic moment and pole strength of each piece respectively ________ and ________.

  1. \(\vec{m}, \frac{p}{2}\)
  2. \(\frac{\vec{m}}{2}, p\)
  3. \(\frac{\vec{m}}{2}, \frac{p}{2}\)
  4. \(\vec{m}, p\)

Answer: 2. \(\frac{\vec{m}}{2}, p\)

Question 7. For superconductors, μr = _________.

  1. Zero
  2. Infinite
  3. Positive
  4. Negative

Answer: 1. Zero

Question 8. Which one of the following represents Curie’s law?

  1. \(M=\frac{C B_0}{T}\)
  2. \(M=\frac{C \chi}{T}\)
  3. \(M=\frac{C \chi}{T-T_C}\)
  4. \(M=\frac{C T}{B_0}\)

Answer: 1. \(M=\frac{C B_0}{T}\)

Question 9. Meissner effect is observed in _________ substance.

  1. Paramagnetic
  2. Ferromagnetic
  3. Superconducting
  4. Permanent magnetic

Answer: 3. Superconducting

Question 10. According to Gauss’s law for magnetism, the net magnetic flux through any closed surface is _________.

  1. Equal to ε0
  2. Infinite
  3. Zero
  4. Equal to μ0

Answer: 3. Zero

Magnetism And Matter Assertion And Reason

For question numbers 1 to 4 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), and (2). (3) and (4) as given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is NOT the correct explanation of A
  3. A is true but R is false
  4. A is false and R is also false

Question 1. Assertion: A bar magnet experiences a torque when placed in a magnetic field.

Reason: A bar magnet exerts a torque on itself due to its magnetic field.

Answer: 3. A is true but R is false

Question 2. Assertion: The magnetic moment of the helium atom is zero.

Reason: All the electrons are paired in helium atom orbitals.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: An atom behaves as a magnetic dipole.

Reason: It is because an atom contains equal positive and negative charges.

Answer: 3. A is true but R is false

Question 4. Assertion: When the radius of a circular loop carrying current is doubled, its magnetic moment becomes four times.

Reason: Magnetic moment depends on the area of the loop.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Magnetism And Matter Short Questions And Answers

Question 1. The magnetic moment of a circular coil carrying current I, having N turns, each of radius r, is M. Find the magnetic moment of the same coil if it is unwound and rewound into a coil having 2N turns for the same current.

Answer:

M = NIA = NIπ2

According to question N(2πr) = 2N(2πr’)

⇒ \(r^{\prime}=\frac{r}{2}\)

So, New magnetic movement M’ = \((2 \mathrm{~N}) \mathrm{I} \pi\left(\frac{\mathrm{r}}{2}\right)^2=\frac{\mathrm{NI} \pi \mathrm{r}^2}{2}=\frac{\mathrm{M}}{2}\)

Question 2. The bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate

  1. The work done in turning the magnet to align its magnetic moment
    1. Normal to the magnetic field,
    2. Opposite to the magnetic field, and
  2. The torque on the magnet in the final orientation in case (2)

Answer:

Magnetic moment M = 6J/T

Magnetic field B = 0.44 T

(1). Work done in rotating the magnet normal to the magnetic field

W = MB(cosθ1 – cosθ2)

where θ1 =60° and θ2 = 90°

W = 6 x 0.44 (cos60°- cos90°) = 6 x 0.44 (0.5- 0) = 1 .32 J

(2). Work done in rotating the magnet opposite to the magnetic field

W = MB(cosθ1 – cosθ2)

Where θ1 =60°and θ2 = 180°

W = 6 x 0.44 (cos60°- cos 180°) = 6 x 0.44 (0.5 + 1 ) = 3.96 J

2. Torque in case (2) τ = MB sin θ2 ⇒ τ = 6 x 0.44 x sin 180° = 0

Question 3.

  1. An iron ring of relative permeability |.ir has windings of insulated copper wire of n turns per meter. When the current in the windings is I, find the expression for the magnetic field in the ring.
  2. The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern by keeping a piece of this material in a uniform magnetic field.

Answer:

1. Consider a ring of radius r having n turns per meter, If n is the number of turns per meter, then the total number of turns in the ring = perimeter times the number of turns = 2πrn = N

Current enclosed = NI = 2πrn I.

By Ampere’s circuital law, \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\)

B x 2πr = μ0 2π rnI or B= μ0nl

Magnetism And Matter Ampere's Circuital Law

2. Given, susceptibility, χ = 0.9853.

The material is paramagnetic.

If a piece of this material is kept in a uniform magnetic field, then the field pattern gets modified as follows:

The lines of force tend to pass through the material rather than the surrounding air.

Magnetism And Matter Paramagnetic In Nature

Question 4. Different diamagnetic, paramagnetic, and ferromagnetic substances based on their properties.

Answer:

Magnetism And Matter Differentiate Diamagnetic, Paramagnetic And Ferromagnetic substances And Their Properties

Magnetism And Matter Long Questions And Answers

Question 1. When the atomic dipoles are aligned partially or fully, there is a net magnetic moment in the direction of the field in any small volume of the material. The actual magnetic field inside the material placed in a magnetic field is the sum of the applied magnetic field and the magnetic field due to magnetization. This field is called magnetic intensity (H).

H = B/μ0-M

where M is the magnetization of the material p0, is the permittivity of vacuum and B is the total magnetic field. The measure that tells us how a magnetic material responds to an external field is given by a dimensionless quantity is appropriately called the magnetic susceptibility; for a certain class of magnetic materials, the intensity of magnetization is directly proportional to the magnetic intensity.

(1). Identify the wrongly matched quantity and unit pair.

  1. Pole strength – A-m
  2. Magnetic susceptibility – dimensionless number
  3. Intensity of magnetisation – A/m
  4. Magnetic permeability- henry-m

Answer: 4. Magnetic permeability- henry-m

(2). A bar magnet has a length of 3 cm, a cross-sectional area of 2 cm2, and a magnetic moment of 3 A-m2. The intensity of magnetization of a bar magnet is

  1. 2 x 105 A/m
  2. 3 x 105 A/m
  3. 4 x 105 A/m
  4. 5 x 105 A/m

Answer: 4. 5 x 105 A/m

(3). A solenoid has a core of material with a relative permeability of 500 and its windings carry a current of A. The number of turns in the solenoid is 500 per meter. The magnetization of the material is nearly:-

  1. 2.5 x 103 A/m
  2. 2.5 x 105 A/m
  3. 2 x 103 A/m
  4. 2 x 105 A/m

Answer: 2. 2.5 x 105 A/m

(4). The relative permeability of iron is 6000. Its magnetic susceptibility is

  1. 5999
  2. 6001
  3. 6000 x 10-7
  4. 6000 x 10-7

Answer: 1. 5999

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