Magnetism & Matter - Live Session - 23 August 2020Contact Number: 9667591930 / 8527521718

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The vertical component of the earths magnetic field is zero at

(1) Magnetic poles

(2) Magnetic equator

(3) Geographic poles

(4) Geographic equator

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

(1) ${\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)$

(2) ${\mathrm{sin}}^{-1}\left(\frac{3}{4}\right)$

(3) ${\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)$

(4) ${\mathrm{sin}}^{-1}\left(\frac{3}{5}\right)$

At a certain place, the horizontal component B_{0} and the vertical component V_{0} of the earth's magnetic field are equal in magnitude. The total intensity at the place will be

1. ${B}_{0}$ 2. ${B}_{0}^{2}$

3. $2{B}_{0}$ 4. $\sqrt{2}{B}_{0}$

The angle of dip at a certain place is 30^{o}. If the horizontal component of the earth’s magnetic field is *H*, the intensity of the total magnetic field is

1. $\frac{H}{2}$ 2. $\frac{2H}{\sqrt{3}}$

3. $H\sqrt{2}$ 4. $H\sqrt{3}$

Time period for a magnet is *T*. If it is divided in four equal parts along its axis and perpendicular to its axis as shown then time period for each part will be

1. 4T

2. T/4

3. T/2

4. T

If a magnetic needle is made to vibrate in uniform field *H*, then its time period is *T*. If it vibrates in the field of intensity 4*H*, its time period will be:

1. 2T

2. T/2

3. 2/T

4. T

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 *oscillations per minute *at a place where dip angle is 30^{o} and 15 *oscillations per minute *at a place where dip angle is 60^{o}. The ratio of total earth's magnetic field at the two places is:

1. $3\sqrt{3}:8$

2. $16:9\sqrt{3}$

3. 4:9

4. $2\sqrt{3}:9$

A magnet of magnetic moment *M* oscillating freely in earth's horizontal magnetic field makes *n* oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled, the number of oscillations made per minute would be

1. $\frac{n}{2\sqrt{2}}$ 2. $\frac{n}{\sqrt{2}}$

3. $2\sqrt{2}n$ 4. $\sqrt{2}n$

Magnets *A *and *B* are geometrically similar but the magnetic moment of *A* is twice that of *B*. If *T*_{1 }and T_{2} be the time periods of the oscillation when their like poles and unlike poles are kept together respectively, then $\frac{{T}_{1}}{{T}_{2}}$ will be:

1. 1/3

2. 1/2

3. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{3}$}\right.$

4. $\sqrt{3}$

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. By doing this the periodic time of the magnetometer will

1. Increase by 19% 2. Decrease by 19%

3. Increase by 11% 4. Decrease by 21%

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

(a) $\frac{4\sqrt{3}}{7}$ (b) $\frac{4}{9\sqrt{3}}$

(c) $\frac{9}{4\sqrt{3}}$ (d) $\frac{9}{\sqrt{3}}$

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

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

(2)${\mathrm{tan}}^{-1}\left(2\right)$

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

(4)None of these

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}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ *seconds*. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is:

1. ${2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

${\mathrm{2.\; 2}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

3. 2

4. ${2}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

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

1. ${\mathrm{cos}}^{2}\varphi ={\mathrm{cos}}^{2}{\varphi}_{1}+{\mathrm{cos}}^{2}{\varphi}_{2}$

2. ${\mathrm{sec}}^{2}\varphi ={\mathrm{sec}}^{2}{\varphi}_{1}+{\mathrm{sec}}^{2}{\varphi}_{2}$

3. ${\mathrm{tan}}^{2}\varphi ={\mathrm{tan}}^{2}{\varphi}_{1}+{\mathrm{tan}}^{2}{\varphi}_{2}$

4. ${\mathrm{cot}}^{2}\varphi ={\mathrm{cot}}^{2}{\varphi}_{1}+{\mathrm{cot}}^{2}{\varphi}_{2}$

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be 2 *seconds*. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be 2 *seconds*. Then the angle of dip is

1. 0^{o} 2. 30^{o}

3. 45^{o} 4. 90^{o}

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2s in earth's horizontal magnetic field of 24 $\mu $T. When a horizontal field is 18 $\mu $T is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be

(a) 1s (b) 2s

(c) 3s (d) 4s

If H is the horizontal component of the earth's magnetic field and 'V' is the vertical component of the earth's magnetic field, then at the magnetic equator:

1. V and H are equal.

2. values of V and H are zero.

3. value of H is zero only.

4. value of V is zero only.

An electric cable is carrying current from north to south. The position of the null point from the cable is:

1. Vertically upward

2. Vertically downward

3. Eastward

4. Nowhere

A magnet (primary) oscillates with frequency ${\mathrm{\nu}}_{1}$ in earth's magnetic field (B_{H}) alone. Now a secondary magnet is placed near the primary magnet. If B is the magnetic field of the second magnet on the primary magnet in the direction of B_{H}, the primary magnet oscillates with frequency ${\mathrm{\nu}}_{2}$, then the ratio of B/B_{H} is :

1. ${\left(\frac{{\mathrm{\nu}}_{2}}{{\mathrm{\nu}}_{1}}\right)}^{2}$

2. ${\left(\frac{{v}_{2}}{{v}_{1}}\right)}^{2}$-1

3. ${\left(\frac{{\mathrm{\nu}}_{2}}{{\mathrm{\nu}}_{1}}\right)}^{2}$ + 1

4. ${\left(\frac{{\mathrm{\nu}}_{1}}{{\mathrm{\nu}}_{2}}\right)}^{2}$-1

If at any place, the angle of dip is $\theta $ and magnetic latitude is $\lambda $, then (assume that the axis of earth's magnetic moment coincides with axis of rotation of the earth)

1. $\theta $ = $\lambda $

2. tan$\theta $ = cos$\lambda $

3. tan$\theta $.cot$\lambda $ = 2

4. tan$\theta $.cot$\lambda $ =$\frac{1}{2}$

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