If a magnet is hanged with its magnetic axis then it stops in

The correct Option is "Magnetic meridian"

Needles ${N_1},\;{N_2}$ and ${N_3}$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

Ferromagnetic substance have strong tendency to get magnetized (induced magnetic moment) in the same direction as that of applied magnetic field, so magnet attract ${N_1}$ strongly. Paramagnetic substances get weakly magnetized (magnetic moment induced is small) in the same direction as that of applied magnetic field, so magnet attracts ${N_2}$ weakly. Diamagnetic substances also get weakly magnatised when placed in an external magnetic field but in opposite direction and hence, ${N_3},$ is weakly repelled by magnet.

Attract ${N_1}\;{\rm{and}}\;{N_2}$ strongly but reple ${N_3}$

The correct Option is "Attract ${N_1}$ strongly, ${N_2}$ weakly and reple ${N_3}$ weakly"

Attract ${N_1}$ strongly, but reple ${N_2}$ and ${N_3}$ weakly

At a certain place, the angle of dip is 30$^ \circ $ and the horizontal component of earth’s magnetic field is 0.50 oersted. The earth’s total magnetic filed (in oersted) is

$R = \frac{H}{{{\rm{cos}}\delta }} = \frac{{0.50}}{{{\rm{cos}}30^\circ }} = \frac{{0.50 \times 2}}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

The correct Option is "$\frac{1}{{\sqrt 3 }}$"

A vibration magnetometer is placed at south pole, then the time period will be

At south pole, $H = 0$ $T = 2\pi \sqrt {\frac{1}{{MH}} \;}= \infty $

The correct Option is "Infinity"

Same as at any other place on earth

The earth’s magnetic field at a certain place has a horizontal component $0.3\;gauss$ and the total strength $0.5\;gauss$. The angle of dip is

${B^2} = B_V^2 + B_H^2 \Rightarrow {B_V} = \sqrt {{B^2} - B_H^2} = \sqrt {{{\left( {0.5} \right)}^2} - {{\left( {0.3} \right)}^2}} = 0.4$ Now $\tan \phi = \frac{{{B_V}}}{{{B_H}}} = \frac{{0.4}}{{0.3}} = \frac{4}{3} \Rightarrow \phi = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$

The correct Option is "${\tan ^{ - 1}}\frac{4}{3}$"

A coil in the shape of an equilateral triangle of side $0.02\;m$ is suspended from its vertex such that it is hanging in a vertical plane between the pole pieces of permanent magnet producing a uniform field of $5 \times {10^{ - 2}}T$. If a current of $0.1\;A$ is passed through the coil, what is the couple acting

Torque $\tau = iAB\sin \theta ,\;i = 0.1A,\;\theta = 90^\circ $ $A = \frac{1}{2} \times base \times height$ or $A = \frac{1}{2}a \times \frac{{a\sqrt 3 }}{2}$ $ = \frac{{\sqrt 3 {a^2}}}{4} = \frac{{\sqrt 3 \times {{\left( {0.02} \right)}^2}}}{4}$ $ = \sqrt 3 \times {10^{ - 4}}{m^2};\;\theta = 90^\circ $ $\tau = 0.1 \times \sqrt 3 \times {10^{ - 4}} \times 5 \times {10^{ - 2}}\sin 90^\circ $ $ = 5\sqrt 3 \times {10^{ - 7}}N - m$

The correct Option is "$5\sqrt 3 \times {10^{ - 7}}N - m$"

$5\sqrt 3 \times {10^{ - 10}}N - m$

$\frac{{\sqrt 3 }}{5} \times {10^{ - 7}}N - m$

Two bar magnets of the same mass, same length and breadth but having magnetic moments $M$ and 3M are joined together pole for pole and suspended by a string. The time period of assembly in a magnetic field of strength $H$ is 3 s. If now the polarity of one of the magnets is reversed and the combination is again made to oscillate in the same field, the time of oscillation is

In pole for pole combination, ${M_1} = 3M\;{\rm{and}}\;{M_2} = M$ ${T_2} = ?$ $\frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{M_1}}}{{{M_2}}}} = \sqrt {\frac{{3\;M}}{M}} = \sqrt 3 $ ${T_2} = \sqrt 3 {T_1} = 3\sqrt 3 $ s

The correct Option is "$3\sqrt 3 $ s"

A magnet of magnetic moment $M$ is rotated through $360^\circ $ in a magnetic field $H$. The work done will be

$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right);\;{\theta _1} = 0^\circ $ and ${\theta _2} = 360^\circ \Rightarrow W = 0$

A coil of 50 turns and area $1.25 \times {10^{ - 3}}\;{{\rm{m}}^2}$ is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 2 A. When the coil is held with its plane is $N - S$ of 2A. When the coil is held with its plane in $N - S$ direction, it experience a couple of 0.04 N-m; and when its plane is $E - W,$ the corresponding couple is 0.03 N-m. The magnetic induction is

$M = ni\;A = 50 \times 2 \times 1.25 \times {10^{ - 3}} = 0.125{\rm{\;A}}{{\rm{m}}^2}$ If normal to the face of the coil makes an angle ${\rm{\theta }}$ with the magnetic induction $B$, then in 1st case, torque $ = MB\;{\rm{cos\theta }} = 0.04$, and in second case, Torque $ = MB{\rm{sin\theta }} = 0.03$ $\therefore MB = \sqrt {{{\left( {0.04} \right)}^2} + {{\left( {0.03} \right)}^2}} = 0.05$ $B = \frac{{0.05}}{M} = \frac{{0.05}}{{0.125}} = 0.4$T

A short bar magnet has a length $2l\;$ and a magnetic moment $10\;{\rm{A}}{{\rm{m}}^2}$. Find the magnetic field at a distance of ${\rm{z}} = 0.1{\rm{\;m}}$ from its centre on the axial line. Here, $l\;$ is negligible as compared to $z$.

On axial position ${B_a} = \frac{{{{\rm{\mu }}_0}}}{{4\pi }}\frac{{2Mr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}};{\rm{if\;}}l < \; < r,\;{\rm{then\;}}{B_a} = \frac{{{{\rm{\mu }}_0}}}{{4\pi }}\frac{{2M}}{{{r^3}}}$ ${B_a} = \frac{{{{10}^{ - 7}}\; \times \;2\; \times \;10}}{{{{\left( {0.1} \right)}^3}}} = 2 \times {10^{ - 3}}\;{\rm{T\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}\left[ {{ \because }z = r} \right]$