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

At a certain place, the horizontal component B₀ and the vertical component V₀ 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. $2B_0$                          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 4H, 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 T₁ and T₂ 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)${\tan }^{-1}\left(\frac{1}{2}\right)$
(2)${\tan }^{-1}\left(2\right)$
(3)${\tan }^{-1}\left(\frac{2}{3}\right)$
(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. ${\cos }^2\varphi ={\cos }^2{\varphi }_1+{\cos }^2{\varphi }_2$
2. ${\sec }^2\varphi ={\sec }^2{\varphi }_1+{\sec }^2{\varphi }_2$
3. ${\tan }^2\varphi ={\tan }^2{\varphi }_1+{\tan }^2{\varphi }_2$
4. ${\cot }^2\varphi ={\cot }^2{\varphi }_1+{\cot }^2{\varphi }_2$

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$\lambda$ =$\frac{1}{2}$