101.
A current carrying coil is placed with its axis perpendicular to $$N-S$$ direction. Let horizontal component of earth’s magnetic field be $${H_0}$$ and magnetic field inside the loop be $$H.$$ If a magnet is suspended inside the loop, it makes angle $$\theta $$ with $$H.$$ Then $$\theta = $$
A
$${\tan ^{ - 1}}\left( {\frac{{{H_0}}}{H}} \right)$$
B
$${\tan ^{ - 1}}\left( {\frac{H}{{{H_0}}}} \right)$$
C
$${\text{cose}}{{\text{c}}^{ - 1}}\left( {\frac{H}{{{H_0}}}} \right)$$
D
$${\cot ^{ - 1}}\left( {\frac{{{H_0}}}{H}} \right)$$
102.
The figure below shows the north and south poles of a permanent magnet in which n turn coil of area of cross-section $$A$$ is resting, such that for a current $$i$$ passed through the coil, the plane of the coil makes an angle $$\theta $$ with respect to the direction of magnetic field $$B.$$ If the plane of the magnetic field and the coil are horizontal and vertical respectively, the torque on the coil will be
A
$$\tau = niAB\cos \theta $$
B
$$\tau = niAB\sin \theta $$
C
$$\tau = niAB$$
D
None of the above, since the magnetic field is radial
Plane of coil is having angle $$\theta $$ with the magnetic field.
$$\therefore \tau = MB\sin \left( {{{90}^ \circ } - \theta } \right)\,{\text{or}}\,\tau = niAB\cos \theta \left( {{\text{As}}\,M = niA} \right)$$
103.
Two short magnets with their axes horizontal and perpendicular to the magnetic maridian are placed with their centres $$40\,cm$$ east and $$50\,cm$$ west of magnetic needle. If the needle remains undeflected, the ratio of their magnetic moments $${M_1}:{M_2}$$ is
104.
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by $${\mu _d},{\mu _p}$$ and $${\mu _f}$$ respectively, then
A
$${\mu _d} \ne 0\,{\text{and}}\,{\mu _f} \ne 0$$
In diamagnetic substances, in each pair of electrons, the spin of both the electrons are in opposite directions. Hence, the electrons of each pair completely cancel the magnetic moment of each other. Thus, the net magnetic moment of each atom of such substances is zero, i.e. $${\mu _d} = 0.$$
The property of para-magnetism is found in those substances whose atoms or molecules have an excess of electrons spinning in same direction.
Hence, atoms of paramagnetic substances have permanent magnetic moment i.e. $${\mu _p} \ne 0$$
The property of ferromagnetism is found in substances which acquire very strong magnetism when placed in an external magnetic field. Like the paramagnetic substances each atom of ferromagnetic substances also has a permanent magnetic moment i.e. $${\mu _f} \ne 0.$$
105.
A very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained $$20\,cm$$ away from the centre of the magnet. If the earth’s magnetic field (horizontal component) at this point be $$0.3\,gauss,$$ the magnetic moment of the magnet is
106.
Two bar magnets having same geometry with magnetic moments $$M$$ and $$2M,$$ are firstly placed in such a way that their similar poles are on the same side, then its period of oscillation is $${T_1}.$$ Now, the polarity of one of the magnets is reversed the time period of oscillations becomes $${T_2}.$$ Then,
The time period of bar magnet is $$T = 2\pi \sqrt {\frac{I}{{MH}}} $$
where, $$M =$$ magnetic moment of magnet
$$I =$$ moment of inertia
and $$H =$$ horizontal component of magnetic field
When same poles of magnets are placed on same side, then net magnetic moment
$$\eqalign{
& {M_1} = M + 2M = 3M \cr
& \therefore {T_1} = 2\pi \sqrt {\frac{I}{{{M_1}H}}} \cr
& = 2\pi \sqrt {\frac{I}{{3MH}}} \,......\left( {\text{i}} \right) \cr} $$
When opposite poles of magnets are placed on same side, then net magnetic moment
$$\eqalign{
& {M_2} = 2M - M = M \cr
& \therefore {T_2} = 2\pi \sqrt {\frac{I}{{{M_2}H}}} = 2\pi \sqrt {\frac{I}{{MH}}} \,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), we observe that
$${T_1} < {T_2}$$