# NCERT Exemplar Solutions for CBSE Class 12 Physics ‒ Chapter 5: Magnetism and Matter (MCQ II)

NCERT Exemplar Solutions for Class12 Physics – Chapter 5 (Moving Charges & Magnetism) are available here. Here, you will get solutions of multiple choice questions with multiple correct answers (i.e., question number 5.6 to question number 5.10). These questions are important for various engineering and medical entrance exams.  These questions can also be asked in CBSE Class 12 Physics board exam.

NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 5 (from question number 5.6 to 5.10) are given below:

Question 5.6: S is the surface of a lump of magnetic material.

(a) Lines of B are necessarily continuous across S.

(b) Some lines of B must be discontinuous across S.

(c) Lines of H are necessarily continuous across S.

(d) Lines of H cannot all be continuous across S.

Solution 5.6: (a), (d)

We know that, magnetic field lines form continuous loops so, lines of B are necessarily continuous across S.

Now, H = B/μo outside the material and H = B/ μo μr inside the material. So, H cannot all be continuous across S.

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Question 5.7: The primary origin(s) of magnetism lies in

(a) atomic currents.

(b) Pauli exclusion principle.

(c) polar nature of molecules.

(d) intrinsic spin of electron.

Solution 5.7: (a), (d)

Magnetism arises due to the spinning electrons revolving around nucleus of an atom.

Question 5.8: A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of μr = 1000. The core is heated beyond the Curie temperature, Tc.

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.

(b) The H and B fields in the solenoid are nearly unchanged.

(c) The magnetisation in the core reverses direction.

(d) The magnetisation in the core diminishes by a factor of about 108.

Solution 5.8: (a), (d)

In a solenoid, H = n I, and B = m0mr nI. So, we can say H field in the solenoid is (nearly) unchanged but the B field decreases drastically

(χm)Fero /m)para » 103/10-5 = 108

The magnetisation in the core diminishes by a factor of about 108.

Question 5.9: Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to

(a) electrostatic field lines can end on charges and conductors have free charges.

(b) lines of B can also end but conductors cannot end them.

(c) lines of B cannot end on any material and perfect shielding is not possible.

(d) shells of high permeability materials can be used to divert lines of B from the interior region.

Solution 5.9: (a), (c), (d)

Electrostatic field lines can end on charges and conductors have free charges.

Lines of B cannot end on any material and perfect shielding is not possible.

Shells of high permeability materials can be used to divert lines of B from the interior region.

Question 5.10: Let the magnetic field on earth be modelled by that of a point magnetic dipole at the centre of earth. The angle of dip at a point on the geographical equator

(a) is always zero.

(b) can be zero at specific points.

(c) can be positive or negative.

(d) is bounded.

Solution 5.10: (b), (c), (d)

The angle of dip at a point on the geographical equator can be zero at specific points. It can be positive or negative. Also, it is bounded in a range from positive to negative value.

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