31.
At a place on earth, horizontal component of earth’s magnetic field is $${B_1}$$ and vertical component of earth’s magnetic field is $${B_2}.$$ If a magnetic needle is kept vertical, in a plane making angle $$\alpha $$ with the horizontal component of magnetic field, then square of time period of oscillation of needle when slightly distributed is proportional to
A
$$\frac{1}{{\sqrt {{B_1}\cos \alpha } }}$$
B
$$\frac{1}{{\sqrt {{B_2}} }}$$
C
$$\frac{1}{{\sqrt {{{\left( {{B_1}\cos \alpha } \right)}^2} + B_2^2} }}$$
Resultant magnetic field in the plane,
$$B = \sqrt {{{\left( {{B_1}\cos \alpha } \right)}^2} + B_2^2} $$
Time period,
$$T = 2\pi \sqrt {\frac{I}{{MB}}} $$
32.
At a temperature of $${30^ \circ }C,$$ the susceptibility of a ferromagnetic material is found to be $$\chi .$$ Its susceptibility at $${333^ \circ }C$$ is
When diamagnetic substances are placed in magnetic field of a strong magnet, then it is feebly magnetised in the opposite direction of field or it is repelled by strong magnet.
34.
A short bar magnet of magnetic moment $$0.4\,J{T^{ - 1}}$$ is placed in a uniform magnetic field of $$0.16\,T.$$ The magnet is in stable equilibrium when the potential energy is
35.
A dip needle lies initially in the magnetic meridian when it shows an angle of dip $$\theta $$ at a place. The dip circle is rotated through an angle $$x$$ in the horizontal plane and then it shows an angle of dip $$\theta '.$$
Then $$\frac{{\tan \theta '}}{{\tan \theta }}$$ is
36.
A Rowland ring of mean radius $$15\,cm$$ has $$3500$$ turns of wire wound on a ferromagnetic core of relative permeability $$800.$$ What is the magnetic field $$B$$ in the core for a magnetising current of $$1.2\,amp$$ ?
Where $$\mu $$ is the permeability of the core and $${\mu _r}$$ is the relative permeability
Now, $${\mu _0} = 4\pi \times {10^{ - 7}}TM{A^{ - 1}},{\mu _r} = 800,I = 1.2\,amp$$
$$\eqalign{
& {\text{and}}\,n = \frac{N}{{2\pi r}} = \frac{{3500}}{{2\pi r}} = \frac{{3500}}{{2\pi \times 0.15}}{m^{ - 1}} \cr
& B = \mu nI = {\mu _0}{\mu _r}nI \cr
& \therefore B = \frac{{\left( {4\pi \times {{10}^{ - 7}}} \right) \times 800 \times 3500 \times 1.2}}{{\left( {2\pi \times 0.15} \right)}} = 4.48\,T \cr} $$
37.
A magnetising field of $$2 \times {10^3}\,amp/m$$ produces a magnetic flux density of $$8\pi \,Tesla$$ in an iron rod. The relative permeability of the rod will be -
38.
The true value of angle of dip at a place is $${60^ \circ },$$ the apparent dip in a plane inclined at an angle of $${30^ \circ }$$ with magnetic meridian is
A
$${\tan ^{ - 1}}\frac{1}{2}$$
B
$${\tan ^{ - 1}}\left( 2 \right)$$
C
$${\tan ^{ - 1}}\left( {\frac{2}{3}} \right)$$
D
$${\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{4}} \right)$$
39.
The magnetic field of earth at the equator is approximately $$4 \times {10^{ - 5}}T.$$ The radius of earth is $$6.4 \times {10^6}m.$$ Then the dipole moment of the earth will be nearly of the order of :
40.
A $$25\,cm$$ long solenoid has radius $$2\,cm$$ and 500 total number of turns. It carries a current of $$15\,A.$$ If it is equivalent to a magnet of the same size and magnetization $$\vec M\left( {\frac{{{\text{magnetic moment}}}}{{{\text{volume}}}}} \right),$$ then $$\left| {\vec M} \right|$$ is: