# Magnetic Dip

## Trending Questions

**Q.**At a certain place, the horizontal component of earth's magnetic field is √3 times the vertical component. The angle of dip at that place is

- 60∘
- 90∘
- 30∘
- 45∘

**Q.**

If ${\theta}_{1}$and ${\theta}_{2}$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip$\left(\theta \right)$ is given by ________

${\mathrm{tan}}^{2}\theta ={\mathrm{tan}}^{2}{\theta}_{1}+{\mathrm{tan}}^{2}{\theta}_{2}$

$co{t}^{2}\theta =co{t}^{2}{\theta}_{1}-co{t}^{2}{\theta}_{2}$

${\mathrm{tan}}^{2}\theta ={\mathrm{tan}}^{2}{\theta}_{1}-{\mathrm{tan}}^{2}{\theta}_{2}$

$co{t}^{2}\theta =co{t}^{2}{\theta}_{1}+co{t}^{2}{\theta}_{2}$

**Q.**The vertical component of earth's magnetic field is zero at or The earth's magnetic field always has a vertical component except at the

- Geographical poles
- Every place
- Magnetic equator
- Magnetic poles

**Q.**If the angles of dip at two places are 30∘ and 45∘ respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be

- √3:√2
- 1:√2
- 1:√3
- 1:2

**Q.**

A magnetic needle is free to rotate in a vertical plane which makes an angle of 600 with the magnetic meridian. If the needle stays in a direction making an angle of tan −1(2√3) with the horizontal, what would be the dip at that place ?

**Q.**

The magnetic needle of a tangent galvanometer is deflected at an angle $30$ due to a magnet. The horizontal component of the earths magnetic field$0.34\times {10}^{-4}T$ is along the plane of the coil. The magnetic intensity is:

**Q.**

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45\xb0$, and $40$ times per minute where the dip is $30\xba$. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B_{1}/ B_{2} is best given by:

$1.8$

$2.2$

$0.7$

$3.6$

**Q.**

What are the magnetic elements of the earths magnetic field?

**Q.**At a place, if the earth’s horizontal and vertical components of magnetic fields are equal, then the angle of dip (in degrees) will be

**Q.**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 tan−1(xy). Find (x+y).

**Q.**

The angle of dip at a place is $\delta $. If the dip is measured in a plane making an angle $\theta $ with the magnetic meridian, the apparent angle of dip ${\delta}_{1}$ will be :

**Q.**

The needle of a dip circle shows an apparent dip of 450 in a particular position and 530 when the circle is rotated through 900. Find the true dip.

**Q.**A dip circle is so set that its needle moves freely in the magnetic meridian. In this position, the angle of dip is 40o. now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30o with the magnetic meridian. In this position, the needle will dip by an angle.

- 30o
- 40o
- more than 40o
- less than 40o

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

- 0∘
- 30∘
- 45∘
- 90∘

**Q.**An electric dipole of length 10 cm having charges 6×10−3 C is placed at 60∘ with respect to a uniform electric field. If it experiences a torque of 3√3 N-m, then the magnitude of electric field is

- 3×104 N/C
- 4×104 N/C
- 1×104 N/C
- 2×104 N/C

**Q.**At a location, vertical component of earth’s magnetic field is 1.6×10−5 T. If dip angle at that location is 37∘, then find the magnitude of magnetic field at that point.

**Q.**

What is the angle of dip?

**Q.**The horizontal and vertical components of magnetic field of earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be __________.

- 90o
- 30o
- 0o
- 45o

**Q.**A long straight horizontal cable carries a current of 2.5 A in the direction 10∘ south of west to 10∘ north of east. The magnetic meridian of the place happens to be 10∘ west of the geographic meridian. The earth's magnetic field at the location is 0.33 G, and the angle of dip is zero. Then neutral points lie on a straight line (ignore the thickness of the cable).

- perpendicular to the cable, below the plane of paper at a perpendicular distance of 1.51 cm.
- parallel to the cable at a perpendicular distance of 3.02 cm.
- perpendicular to the cable at a perpendicular distance of 3.02 cm.
- parallel to the cable, above the plane of paper at a perpendicular distance of 1.51 cm.

**Q.**

The magnetic field due to the earth has a horizontal component of 26μ T at a place where the dip is 600. Find the vertical component and the magnitude of the field.

**Q.**At a certain place, the angle of dip 30o and the horizontal component of earth's magnetic field is 0.50 oerested.The earth's total magnetic field (in oerested) is

- 1
- 1√3
- 12
- √3

**Q.**A dip circle lies initially in the magnetic meridian. If it is now rotated through angle θ in the horizontal plane, then tangent of the angle of dip is changed in the ratio

- 1:cosθ
- cosθ:1
- 1:sinθ
- sinθ:1

**Q.**

If we place a bar magnet in the magnetic meridian with its north pole towards geographical North, the neutral point will be

1.on the axial of the magnet

2.on the equatorial line line of magnet

3.at 45 with the axial line

4. At 45 with equatorial line

**Q.**A dip placed in a plane which is perpendicular to magnetic meridian will remain

- Parallel to surface of earth
- At angle 11.2∘ with surface of earth
- At angle 11.2∘ with vertical
- Vertical

**Q.**What is the magnetic field at point P if AP is making an angle 15∘ as shown in the figure.

- μ04πI2a[√3−√2]
- μ02πIa[√3−√2]
- μ0πIa[√3−√2]
- none of these

**Q.**The horizontal component of earth's magnetic field at a place is 1√3 times the vertical component. Determine the angle of dip at that place.

**Q.**A magnetic needle freed to rotate around a fixed vertical plane stay at an angle 600 with the horizontal . If the dip at that place is 370, find the angle of the fixed vertical plane with the meridian.

- θ=cos−1(√34)
- θ=cos−1√35
- θ=cos−123
- θ=cos−11√3

**Q.**If a dip circle is placed in a vertical plane at an angle of 30o– to the magnetic meridian, the dip needle makes an angle of 45o– with the horizontal. The real dip at that place is?

- tan−1(√3/2)
- tan−1(√3)
- tan−1(√3/√2)
- tan−1(2/√3)

**Q.**In a physics lab, we took two readings of dip angle by dip circle, in two readings planes of dip circle are mutually perpendicular, then ratio of vertical & horizontal component of field the to earth in lab is (reading 1 = 450 reading 2 = 600)

- cot−1{43}
- cot−1{√73}
- cot−1{√43}
- cot−1{73}

**Q.**

If a magnet is suspended at an angle 300 to the magnetic meridian, the dip needle makes an angle of 60o with the horizontal. The true value of dip is :

- tan−1(2/3)
- tan−1(3/2)
- tan−1(3)
- tan−1(2)