Class 12-science H C VERMA Solutions Physics Chapter 14 - Permanent Magnets
Permanent Magnets Exercise 277
Solution 1
m=10 A-m
r=5cm=0.05m
Magnetic field, 
T
Solution 2
N
Solution 3
T-m
Solution 4
Perpendicular distance between the equi-potential surfaces
Now,
Magnetic field, 
Direction of magnetic field is from higher potential to lower potential and perpendicular to equi-potential surface.
So, magnetic field is at 
with the positive x-axis.
Solution 5
T
r=10cm=0.1m
(a) For end-on position
M=1 A-
(b) For broadside-on position
M=2 A-
Solution 6
Let at point P, magnetic field is perpendicular to the magnetic axis.
Here, 
We know that,
Solution 7
Length of the magnet, 2l=8cm⟹l=4cm
Here, r=3cm
B=
T
At broadside-on position
A-m
Solution 8
The neutral point for a magnet with north pole pointing in north will be at broadside on position.
At neutral point
Magnetic field due to magnet=earth's horizontal magnetic field
r=0.2m=20cm
from the midpoint of magnet of plane bisecting the dipole.
Solution 9
The neutral point for a magnet with north pole pointing in south will be at end on position.
At neutral point
Magnetic field due to magnet=earth's horizontal magnetic field
r=0.2m=20cm
from center of magnet.
Solution 10
Let at point P, net magnetic field is zero.
Here, 
We know that,
At neutral point,
Magnetic field due to magnet= earth's horizontal field
at an angle of 
south of east.
Solution 11
M=8.0×1022 A-m2
r=6400km
At axial point, magnetic field
Solution 12
We know that,
Solution 13
Now, total magnetic field,
B=
Solution 14
Apparent dip 
Angle between magnetic meridian and
plane of needle 
Let true dip be δ
We know that,
Permanent Magnets Exercise 278
Solution 15
True dip (δ) is given by
Solution 16
T
R=10cm=0.1m
I=10mA
We know that
Solution 17
N-m
Solution 18
Solution 19
r=0.079m=7.9cm from the center
Solution 20
For short magnet,
m=2cm
From the needle, north pole pointing towards south.
Solution 21
M=1600
Solution 22
Time period is given by 
Let magnetic moment of two poles be M1 and M2 and moment of inertia be l1and l2 respectively.
For like poles
For unlike poles
Divide (i) and (ii)
Solution 23
Initially, magnetic field 
and time period T=0.1sec
Now, when wire is placed, net magnetic field
We know that,
sec
Solution 24
Initially, time period of one
oscillation=
min
And moment of inertia=I
Later, moment of inertia=2I
We know that,
min
For 40 oscillations, time period=
minutes
Solution 25
Frequency oscillation 
(a) for north facing north,
oscillation/minute
(b) for north pole facing south
oscillation/minute
