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Two identical short bar magnets, each having magnetic moment M, are placed a distance of 2d apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

$\frac{μ_{0}}{4 π} (\sqrt{2}) \frac{M}{d^{3}}$

$\frac{μ_{0}}{4 π} (\sqrt{3}) \frac{M}{d^{3}}$

$(\frac{2 μ_{0}}{4 π}) \frac{M}{d^{3}}$

$\frac{μ_{0}}{4 π} (\sqrt{5}) \frac{M}{d^{3}}$

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A magnet oscillating in a horizontal plane has a time period of 2 second at a place where the angle of dip is 30^o and 3 seconds at another place where the angle of dip is 60^o. The ratio of resultant magnetic fields at the two places is

$\frac{4 \sqrt{3}}{7}$

$\frac{4}{9 \sqrt{3}}$

$\frac{9}{4 \sqrt{3}}$

$\frac{9}{\sqrt{3}}$

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Explanation

T $α \frac{1}{\sqrt{B_{H}}} = \frac{1}{\sqrt{B \cos \emptyset}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{B_{2} \cos \emptyset_{2}}{B_{1} \cos \emptyset_{1}}}$

$\Rightarrow \frac{B_{1}}{B_{2}} = \frac{{T^{2}}_{2}}{{T^{2}}_{1}} \times \frac{\cos \emptyset 2}{\cos \emptyset 1} = {(\frac{3}{2})}^{2} \times \frac{\cos 60^{\circ}}{\cos 30^{\circ}} \Rightarrow \frac{B_{1}}{B_{2}} = \frac{9}{4 \sqrt{3}}$

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A superconductor exhibits perfect :

Ferrimagnetism

Ferromagnetism

Paramagnetism

Diamagnetism

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Explanation

(d)

A superconducting material kept in a magnetic field expels the magnetic flux out its body when cooled below the critical temperature and exhibits perfect diamagnetism.

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Among the following properties describing diamagnetism identify the property that is wrongly stated

Diamagnetic material do not have permanent magnetic moment

Diamagnetism is explained in terms of electromagnetic induction

Diamagnetic materials have a small positive susceptibility

The magnetic moment of individual electrons neutralize each other

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Explanation

(c)

Susceptibility of diamagnetic substance is negative and it does not change with temperature.

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If a magnet is suspended at an angle 30^o to the magnetic meridian, it makes an angle of 45^owith the horizontal. The real dip is

$\tan^{- 1} (\frac{\sqrt{3}}{2})$

$\tan^{- 1} (\sqrt{3})$

$\tan^{- 1} (\sqrt{\frac{3}{2}})$

$\tan^{- 1} (\frac{2}{\sqrt{3}})$

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The true value of angle of dip at a place is 60^o, the apparent dip in a plane inclined at an angle of 30^o with magnetic meridian is

$\tan^{- 1} (\frac{1}{2})$

$\tan^{- 1} (2)$

$\tan^{- 1} (\frac{2}{3})$

None of these

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A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{\frac{5}{4}}$ seconds. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is :

$2^{\frac{1}{4}}$

$2^{\frac{1}{2}}$

$2^{\frac{3}{4}}$

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A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of 5.30 × 10³Amp/m³. What is its magnetic dipole moment ?

$1 \times 10^{- 2} J / T$

$2.08 \times 10^{- 2} J / T$

$3.08 \times 10^{- 2} J / T$

$1.52 \times 10^{- 2} J / T$

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If $ϕ_{1}$ and $ϕ_{2}$ be the angles of dip observed in two vertical planes at right angles to each other and $ϕ$ be the true angle of dip, then

$\cos^{2} ϕ = \cos^{2} ϕ_{1} + \cos^{2} ϕ_{2}$

$\sec^{2} ϕ = \sec^{2} ϕ_{1} + \sec^{2} ϕ_{2}$

$\tan^{2} ϕ = \tan^{2} ϕ_{1} + \tan^{2} ϕ_{2}$

$\cot^{2} ϕ = \cot^{2} ϕ_{1} + \cot^{2} ϕ_{2}$

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A dip needle lies initially in the magnetic meridian when it shows an angle of dip $θ$ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip $θ'$ . Then $\frac{\tan θ'}{\tan θ}$ is

1/cos x

1/sin x

1/tan x

cos x

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