NCERT Solutions for Class 12th Physics, Chapter 5: Magnetism and Matter are available here. Due to a large number of problems in this chapter, we have provided the solutions in two parts (Part I and Part II). In Part I, solutions from number 5.1 to 5.8 are available. In Part II (or this part) solutions from question number 5.9 to 5.15 are available. You can also download these NCERT Solutions in PDF format.
NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 5: Magnetism and Matter from question number 5.9 to 5.15 are given below
Question 5.9: A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10^{–2} T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^{–1}. What is the moment of inertia of the coil about its axis of rotation?
Solution5.9:
Given,
N = 16
r = 10 cm
I = 0.75 A
B = 0.05 T
θ = 90
f = 2 / s
The moment of inertia is described by the formula
I = (mB/4π^{2}f^{2}) …(1)
Determining the value of magnetic moment, m = NIA
⇒ m = 16 × 0.75 × 3.14 × 0.1 × 0.1 = 0.377 J / T
Substituting the values in eq …(1)
Moment of inertia of coil
I = 1.19 × 10^{‒4} kgm^{2}.
CBSE Class 12th Physics Notes: All Chapters
Question 5.10: A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22^{o} with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Solution5.10:
Given:
Angle of dip, I = 22 degrees
H_{e} = 0.35G
Using the relation,
cos22 = H_{e} / B_{e} ⇒ B_{e} = 0.35G / cos 22 = 0.377 G.
Question 5.11: At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Solution5.11:
Given,
I = 60 degrees
H_{e }= 0.16G
Using the approach of previous question
B_{e} = 0.16G / cos60 = 0.32G
Direction: magnetic north to magnetic south.
Question 5.12: A short bar magnet has a magnetic moment of 0.48 J T^{–1}. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on
(a) the axis,
(b) the equatorial lines (normal bisector) of the magnet.
Solution5.12:
Given:
m = 0.48 J / T
r = 10 cm
The magnet is short. This indicates that the length of magnet is insignificant when compared to the distance at which the magnitude and direction of magnetic field is to be determined.
r ≫��
The magnitude of the magnetic field at a distance of 10 cm by magnet will be different for both the cases. The direction in both the cases will be different too.
Calculating for the cases:
(a) On the axis
Magnetic field is defined by the formula
B = (μ_{o}2m) / (4πr^{3})
Substitution values, we have, B = 9.6 x 10^{‒5} T
Direction: South to North
(b) On normal bisector
B = (μ_{o}m) / (4πr^{3})
Solving
B’ = 4.8 x 10^{‒5} T
Direction: North to South
Question 5.13: A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Solution5.13:
B_{e }= 0.36G
I = 0 degree
Null point at r = 14 cm
Since the given magnet is short, the length of the magnet plays no significant role in our calculations. The magnet is aligned along the earth’s magnetic field.
Determining the horizontal component of magnetic field, H_{e} / B_{e} = cos0
⇒ H_{e} / B_{e} = 1 ⇒ H_{e} = B_{e} = 0.36G
To calculate the magnetic field at normal bisector of magnet, we have to determine the magnetic moment of magnet first.
Magnetic field due to magnet at its axis at null point,
B’ = (μ_{o}2m)/(4πr^{3})
This magnetic field will be equal and opposite to horizontal component of earth’s magnetic field at null point.
So,
B’ = H_{e}
⇒ (10^{‒7}).(2m) / (0.14)^{3} = 0.36 × 10^{‒4}
This gives, m = 0.49392 J / T
Now determining the magnetic field at normal bisector of magnet at given distance of r = 14 cm.
B = (μ_{o}m)/(4πr^{3})
Which yields,
B = 0.18 G
Now
Total Magnetic field = magnetic field due to magnet + magnetic field due to earth
Therefore Total Magnetic field = 0.18G + 0.36G = 0.54G
Direction: In the direction of earth’s field.
Question 5.14: If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?
Solution5.14:
New null point at normal bisector
B = (μ_{o}m/4πr^{3})
It will be equal and opposite to horizontal component of the earth’s magnetic field at that point.
We have already determined magnetic moment in the previous question.
Using that value,
(10^{‒7}) × (0.4939) / (0.36 x 10^{‒4}) = r^{3} ⇒ r = 0.11 m.
Thus the null point will be at a distance of r = 11.1 cm on the normal bisector of the given magnet.
Question 5.15: A short bar magnet of magnetic moment 5.25 × 10^{–2} J T^{–1} is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45^{o} with earth’s field on
(a) its normal bisector and
(b) its axis.
Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Solution 5.15:
Given:
m = 5.25 x 10-2 J / T
θ = 90 degrees
B = 0.42 G
(a) At normal bisector
B = (μ_{o}m/4πr^{3})
Putting the values and solving for r, r = 0.05 m or r = 5 cm.
(b) At axis
B = (μ_{o}2m/4πr’^{3})
As solved in previous part, r’ = 0.063 m or r^{’} = 6.3 cm.
Download NCERT Solutions for Class 12 Physics ‒ Chapter 5: Magnetism and Matte in PDF format
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