NCERT Exemplar Solutions for CBSE Class 12 Physics ‒ Chapter 4: Moving Charges & Magnetism (Part II)
NCERT Exemplar Solutions for Class12 Physics – Chapter 4 (Moving Charges & Magnetism) are available here. In this article, you will get solutions of multiple choice questions with multiple correct answers i.e., from question number 4.7 to 4.11. Solutions of questions from 4.1 to 4.6 are already available in Part I. Solutions of other questions will be available in other parts. These questions are important for various engineering and medical entrance exams like JEE Mains, NEET, WBJEE etc.
NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 4 (from question number 4.7 to 4.11) are given below:
Question 4.7: The gyro-magnetic ratio of an electron in an H-atom, according to Bohr model, is
(a) independent of which orbit it is in.
(d) increases with the quantum number n.
The gyro-magnetic ratio of an electron in an H-atom is equal to the ratio of the magnetic moment and the angular momentum of the electron. Both are related as,
μl = ‒ (e/2me) l
The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.
Question 4.8: Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
(a) motion of charges inside the conductor is unaffected by B since they do not absorb energy.
(b) some charges inside the wire move to the surface as a result of B.
(c) if the wire moves under the influence of B, no work is done by the force.
(d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.
Soluion4.8: (b, d)
Magnetic force on a conductor of length l carrying a current I placed in a uniform magnetic field B is given by, F = I (l × B) or |F| = I |l|| B| sin θ.
The direction of force is given by Fleming's left hand rule & F is perpendicular to the direction of magnetic field B, Therefore, work done by the magnetic force on the ions is zero.
Soluion4.9: (b, c)
Question 4.10: A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity - v. At this instant,
(a) the electric forces on both the particles cause identical accelerations.
(b) the magnetic forces on both the particles cause equal accelerations.
(c) both particles gain or loose energy at the same rate.
(d) the motion of the centre of mass (CM) is determined by B alone.
Soluion4.10: (b, c, d)
Question 4.11: A charged particle would continue to move with a constant velocity in a region wherein,
(a) E = 0, B ≠ 0.
(b) E ≠ 0, B ≠ 0.
(c) E ≠ 0, B = 0.
(d) E = 0, B = 0.
Soluion4.11: (a, b, d)
Force on charged particle due to magnetic field, Fm = q(v × B)
Force on charged particle due to electric field FE = qE.
Now, FE = 0 if E = 0
Also, Fm = 0 if sin q = 0 or q° = 0° or 180°.
It may be also possible that resultant of q E + q (v × B) is equal to zero.
Therefore, options (a), (b) and (d) are correct.