NCERT Exemplar Solutions for CBSE Class 12 Physics ‒ Chapter 4: Moving Charges & Magnetism (Part I)
Find NCERT Exemplar Solutions for Class12 Physics – Chapter 4 (Moving Charges & Magnetism). Here, you will get solutions of multiple choice questions with single correct answer (i.e., question number 4.1 to question number 4.6). 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 4 (from question number 4.1 to 4.6) are given below:
Question 4.1: Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B0ˆk.
(a) They have equal z-components of momenta.
(b) They must have equal charges.
(c) They necessarily represent a particle-antiparticle pair.
(d) The charge to mass ratio satisfy: (e/m)1 + (e/m)2 = 0.
Solution 4.1: (d)
When charge/mass ratio of these two particles is same and charges on them are of opposite nature then the charged particles will traverse identical helical paths in a completely opposite sense. Therefore, option (d) is correct.
Question 4.2: Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that
(a) B ⊥ v.
(b) B ||v.
(c) it obeys inverse cube law.
(d) it is along the line joining the electron and point of observation.
Solution 4.2: (a)
In Biot-Savart's law, magnetic field B|| idl × r and idl due to flow of electron is in opposite direction of v and by direction of cross product of two vectors, B ^ v.
Question 4.3: A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane.
(a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0.0.z), z >>R increases.
(d) The magnitude of B at (0.0.z), z >>R is unchanged.
Solution 4.3: (a)
For a circular loop of radius R, carrying current I in x-y plane, the magnetic moment M = I × π R2.
It acts perpendicular to the loop along z-direction.
When half of the current loop is bent in y-z plane, then magnetic moment due to half current loop is x-y plane, M1 = I (πR2/2) acting along z-direction.
Magnetic moment due to half current loop in y-z plane, M2 = I (πR2/2) along x-direction.
Net magnetic moment due to entire bent current loop, Mnet = (M’2 + M’2)1/2 = √2 M’ = √2 I (πr2)/4. Therefore, M net < M or M diminishes.
Question 4.4: An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?
(a) The electron will be accelerated along the axis.
(b) The electron path will be circular about the axis.
(c) The electron will experience a force at 45° to the axis and hence execute a helical path.
(d) The electron will continue to move with uniform velocity along the axis of the solenoid.
Solution 4.4: (d)
The magnetic field inside the long current carrying solenoid is uniform. Therefore, magnitude of force on the electron of charge (‒ e) is given by F = ‒ evB sinθ 180° = 0 (θ = 0o) as magnetic field and velocity are parallel. The electron will continue to move with uniform velocity along the axis of the solenoid.
Question 4.5: In a cyclotron, a charged particle
(a) undergoes acceleration all the time.
(b) speeds up between the dees because of the magnetic field.
(c) speeds up in a dee.
(d) slows down within a dee and speeds up between dees.
Solution 4.5: (a)
In a cyclotron, a charged particle describes the circular path inside the dees & is accelerated while going from one dee to another due to the electric field.
While moving in the circular path, charged particles have centripetal acceleration which is provided by the magnetic force due to the magnetic field.
A charged particle undergoes acceleration all the time, inside the cyclotron.
Question 4.6: A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is
(b) √3 MB/2.
Solution 4.6: (d)
The work done to rotate the loop in magnetic field W = MB (cos θ1 ‒ cos θ2).
When current carrying coil is rotated then there will be no change in angle between magnetic moment and magnetic field.
Here, θ1 = θ2 = α
⇒ W = MB (cos α ‒ cos α) = 0.