NCERT Exemplar Solutions for Class12 Physics – Chapter 5 (Moving Charges & Magnetism) are available here. Here, you will get solutions of multiple choice questions with single correct answer (i.e., question number 5.1 to question number 5.5). 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 5.1 to 5.5) are given below:
A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m
(a) is non-zero and points in the z-direction by symmetry.
(b) points along the axis of the tortoid (m = m φ ).
(c) is zero, otherwise there would be a field falling as 1/r3 at large distances distances outside the toroid.
(d) is pointing radially outwards.
Solution 5.1: (c)
In case of toroid, the magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and there is no magnetic field outside the body of toroid. This is because the loop encloses no current. Thus, the magnetic moment of toroid is zero.
The magnetic field of Earth can be modelled by that of a point dipole placed at the centre of the Earth. The dipole axis makes an angle of 11.3° with the axis of Earth. At Mumbai, declination is nearly zero.
(a) the declination varies between 11.3° W to 11.3° E.
(b) the least declination is 0°.
(c) the plane defined by dipole axis and Earth axis passes through Greenwich.
(d) declination averaged over Earth must be always negative.
Solution 5.2: (a)
The axis of the dipole does not coincide with the axis of rotation of the earth. It makes an angle of 11.3o with the axis of the earth, the declination varies between 11.3o W to 11.3o E depending on the observation point.
Question 5.3: In a permanent magnet at room temperature
(a) magnetic moment of each molecule is zero.
(b) the individual molecules have non-zero magnetic moment which are all perfectly aligned.
(c) domains are partially aligned.
(d) domains are all perfectly aligned.
Solution 5.3: (c)
In case of a permanent magnetic, at room temperature, the magnetic domains are partially aligned.
Solution 5.4: (b)
A paramagnetic sample shows a net magnetisation of 8 Am–1 when placed in an external magnetic field of 0.6T at a temperature of 4K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16K, the magnetisation will be
(a) (32/3) Am–1
(b) (2/3) Am–1
(c) 6 Am–1
(d) 2.4 Am–1
Solution 5.5 : (b)
From the Curie’s law explains, I ∝ B/T
Here, B is magnetic field induction, I is the intensity of magnetisation and T is absolute temperature.
⇒ (I2/I1) = (B2/B1) × (T1/T2)
I1 = 8 Am-1, B1 = 0.6T, T1 = 4 K, B2 = 0.2T, T2 = 16 K, I2 =?
By putting values in equation …(i), we have,
I2 = (2/3) Am‒1.