HC Verma Questions and Solutions: Chapter 36: Permanent Magnets- 1 | HC Verma Solutions - JEE PDF Download

Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position.
If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.

A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in end-on position.

When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength m and the distance between the poles are not fixed.
But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, md is fixed.
A current loop of area A and current I can be replaced with a magnetic dipole of dipole moment md.
i.e. md = IA

Magnetic field B due to a bar magnet of magnetic moment M at distance r of the point on the axis of the magnet from its centre is given by

Here, 2l  is the length of the magnet.
So, from the above formula, it can be easily seen that 

Magnetic field B due to a bar magnet of magnetic moment M at distance r of the point on the axis from its centre is given by 
Here, 2l  is the length of the magnet.
When the distance of the point where the magnetic field has to be calculated is greater than the length of the magnet, i.e r >> l , the bar magnet acts like a magnetic dipole whose magnetic field is B ∝ (1/r³)
Now, l in the denominator can be neglected.
So, the correct option is (c).

Magnetic field (B₁) due to the short dipole A of dipole moment M at an axial point  is given by,  ...(1)
Magnetic field (B₂) due to the short dipole B of dipole moment M at an axial point is given by,  ...(2)
Resultant magnetic field (B) will be,

Magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet.

When taken to a geomagnetic pole, a compass needle that is allowed to move in a horizontal plane will try to suspend itself vertically to the horizontal plane containing the compass. In other words, the horizontal plane containing the compass will restrict the compass to suspend itself in vertical direction; hence, the compass will stay in any position.
However, a freely suspended magnet will become vertical at poles, with its north pole pointing towards Earth at its north pole (which is magnetic south).

At the geomagnetic equator, the needle tries to suspend itself in horizontal direction. But here the needle is restricted to move only in the vertical plane perpendicular to the magnetic meridian. Hence, the needle will stay in the direction it is released.

For a tangent galvanometer, deflection is given by

Here, k is the constant called reduction factor.
From the above formula, we can say that deflection is independent of the number of turns.
Hence, on doubling the number of turns, deflection remains the same.

The current and deflection dependence of a moving coil galvanometer is given by 
Therefore, if we double the current, the deflection also gets doubled.
However, in a tangent galvanometer, i ∝ tan θ;  that is, there is no direct relation between θ and current.
Hence, the correct option is (b).

In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current i. So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.
Let idl be the current element, B be the magnetic field and dF be the force on the current element idl.
Now

Thus, the force acting on the wire is  and it is perpendicular to the plane of the wire.

Investigators and experimenters have failed to find any sign of magnetic monopoles. So, we can assume that magnetic monopoles are only a mathematical assumption.
A magnetic field is produced by the motion of an electric charge only. In paramagnets or ferromagnets, the motion of an electron (charge) and the alignment of domains (bunch of charges with particular alignment) create paramagnetism and ferromagnetism, respectively.
Therefore, the only cause behind the magnetic field is the motion of an electric charge.
Denial of (c):
The north pole is equivalent to an anticlockwise current and the south pole is equivalent to a clockwise current.
Denial of (d):
A bar magnet is not equivalent to a long, straight current because the distribution and orientation of magnetic field lines do not resemble each other.

A horizontal circular loop carrying current in clockwise direction acts like the south pole of a magnet. Hence, the south pole of the magnet coincides with the loop.

Now, when the loop carrying current in clockwise direction is viewed from above, it looks like the magnetic lines of force are entering the loop thus it acts like south pole of a magnet. And if we view from below the loop then it appears that magnetic lines of force are leaving the loop.  Hence, the north pole should be below the loop.

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of magnetic field  is the same only at points P₁ and P₂ and at points Q₁ and Q₂.

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of the magnetic field  is opposite at points P₁ and Q₁ and at points P₂ and Q₂.

Denial of (a):
Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.
Justification of (b) and (c) :
Deflection magnetometer is used to measure M/B_H of a permanent bar magnet.
Similarly, oscillation magnetometer is used to measure MB_H of a bar magnet. So, if earth's horizontal field, B_H is known, then the magnetic moment of a bar magnet, M, can be measured.
Justification of (d):
Using deflection and oscillation magnetometers, we can calculate  respectively . Therefore, if we multiply the result obtained from both the instruments, then B_H cancels out as  Thus, the value of B_H is not required.
Therefore, we can use both deflection and oscillation magnetometers if the earth's horizontal field is not known.