QBManager

PHXII05:MAGNETISM and MATTER

360683 At certain place, horizontal component is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip is

1

30^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

Here, horizontal component
$B_{H} = \frac{1}{\sqrt{3}} \times B_{V}$
$\tan δ = \frac{B_{V}}{B_{H}} = \frac{B_{V}}{\frac{1}{\sqrt{3}} B_{V}} = \sqrt{3}$
$\tan δ = \sqrt{3} \Rightarrow δ = \tan^{- 1} (\sqrt{3}) = 60^{\circ}$ .
So correct option is (4)

PHXII05:MAGNETISM and MATTER

360684 If $B_{H} = \frac{1}{\sqrt{3}} B_{V}$ , then find angle of dip. ( where, symbols have their usual meanings)

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

Magnetic dip or magnetic inclination is given by
$\tan δ = \frac{B_{V}}{B_{H}} (1)$
Where, $B_{V}$ and $B_{H}$ are vertical and horizontal components of earth's magnetic field, respectively.
Given, $B_{H} = \frac{1}{\sqrt{3}} B_{V} \Rightarrow \frac{B_{V}}{B_{H}} = \sqrt{3} (2)$
From Eqs. (1) and (2), we get
$\tan δ = \sqrt{3} \Rightarrow δ = 60^{\circ}$

PHXII05:MAGNETISM and MATTER

360685 Consider the earth as a short magnet with its centre coinciding with the centre of the earth and dipole moment $M$ . The angle of dip $δ$ is related to latitude $λ$ as

1

\tan δ = \frac{\tan λ}{2}

2

\tan δ = 2 \tan λ

3

\tan δ - \cot λ

4

\tan δ = \tan λ

Explanation:

$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos θ, B_{θ} = \frac{μ_{0} M \sin θ}{4 π r^{3}}$
as
$θ = 90 - λ$
$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos (90 - λ)$
$= \frac{2 μ_{0} M \sin λ}{4 π r^{3}}$
$\frac{B_{r}}{B_{θ}} = \frac{B_{r}}{B_{θ}} = 2 \tan λ o r \tan δ = 2 \tan λ$
supporting img

PHXII05:MAGNETISM and MATTER

360686 The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{- 4} W b m^{- 2}$ . If the angle of dip at that place is $60^{\circ}$ , then the value of vertical component of the earth's magnetic field will be (in $W b m^{- 2}$ )

1

0.12 \times 10^{- 4}

2

0.24 \times 10^{- 4}

3

0.40 \times 10^{- 4}

4

0.622 \times 10^{- 4}

Explanation:

Vertical component of earth's magnetic field,
$B_{V} = B_{H} \tan δ$
$= 0.36 \times 10^{- 4} \times \tan 60^{0}$ $(g i v e n)$
$= 0.36 \times 10^{- 4} \times \sqrt{3} = 0.622 \times 10^{- 4} T$
$= 0.622 \times 10^{- 4} W b m^{- 2}$
So, correct option is (4)

AIIMS - 2018

PHXII05:MAGNETISM and MATTER

360683 At certain place, horizontal component is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip is

1

30^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

Here, horizontal component
$B_{H} = \frac{1}{\sqrt{3}} \times B_{V}$
$\tan δ = \frac{B_{V}}{B_{H}} = \frac{B_{V}}{\frac{1}{\sqrt{3}} B_{V}} = \sqrt{3}$
$\tan δ = \sqrt{3} \Rightarrow δ = \tan^{- 1} (\sqrt{3}) = 60^{\circ}$ .
So correct option is (4)

PHXII05:MAGNETISM and MATTER

360684 If $B_{H} = \frac{1}{\sqrt{3}} B_{V}$ , then find angle of dip. ( where, symbols have their usual meanings)

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

Magnetic dip or magnetic inclination is given by
$\tan δ = \frac{B_{V}}{B_{H}} (1)$
Where, $B_{V}$ and $B_{H}$ are vertical and horizontal components of earth's magnetic field, respectively.
Given, $B_{H} = \frac{1}{\sqrt{3}} B_{V} \Rightarrow \frac{B_{V}}{B_{H}} = \sqrt{3} (2)$
From Eqs. (1) and (2), we get
$\tan δ = \sqrt{3} \Rightarrow δ = 60^{\circ}$

PHXII05:MAGNETISM and MATTER

360685 Consider the earth as a short magnet with its centre coinciding with the centre of the earth and dipole moment $M$ . The angle of dip $δ$ is related to latitude $λ$ as

1

\tan δ = \frac{\tan λ}{2}

2

\tan δ = 2 \tan λ

3

\tan δ - \cot λ

4

\tan δ = \tan λ

Explanation:

$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos θ, B_{θ} = \frac{μ_{0} M \sin θ}{4 π r^{3}}$
as
$θ = 90 - λ$
$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos (90 - λ)$
$= \frac{2 μ_{0} M \sin λ}{4 π r^{3}}$
$\frac{B_{r}}{B_{θ}} = \frac{B_{r}}{B_{θ}} = 2 \tan λ o r \tan δ = 2 \tan λ$
supporting img

PHXII05:MAGNETISM and MATTER

360686 The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{- 4} W b m^{- 2}$ . If the angle of dip at that place is $60^{\circ}$ , then the value of vertical component of the earth's magnetic field will be (in $W b m^{- 2}$ )

1

0.12 \times 10^{- 4}

2

0.24 \times 10^{- 4}

3

0.40 \times 10^{- 4}

4

0.622 \times 10^{- 4}

Explanation:

Vertical component of earth's magnetic field,
$B_{V} = B_{H} \tan δ$
$= 0.36 \times 10^{- 4} \times \tan 60^{0}$ $(g i v e n)$
$= 0.36 \times 10^{- 4} \times \sqrt{3} = 0.622 \times 10^{- 4} T$
$= 0.622 \times 10^{- 4} W b m^{- 2}$
So, correct option is (4)

AIIMS - 2018

PHXII05:MAGNETISM and MATTER

360683 At certain place, horizontal component is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip is

1

30^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

Here, horizontal component
$B_{H} = \frac{1}{\sqrt{3}} \times B_{V}$
$\tan δ = \frac{B_{V}}{B_{H}} = \frac{B_{V}}{\frac{1}{\sqrt{3}} B_{V}} = \sqrt{3}$
$\tan δ = \sqrt{3} \Rightarrow δ = \tan^{- 1} (\sqrt{3}) = 60^{\circ}$ .
So correct option is (4)

PHXII05:MAGNETISM and MATTER

360684 If $B_{H} = \frac{1}{\sqrt{3}} B_{V}$ , then find angle of dip. ( where, symbols have their usual meanings)

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

Magnetic dip or magnetic inclination is given by
$\tan δ = \frac{B_{V}}{B_{H}} (1)$
Where, $B_{V}$ and $B_{H}$ are vertical and horizontal components of earth's magnetic field, respectively.
Given, $B_{H} = \frac{1}{\sqrt{3}} B_{V} \Rightarrow \frac{B_{V}}{B_{H}} = \sqrt{3} (2)$
From Eqs. (1) and (2), we get
$\tan δ = \sqrt{3} \Rightarrow δ = 60^{\circ}$

PHXII05:MAGNETISM and MATTER

360685 Consider the earth as a short magnet with its centre coinciding with the centre of the earth and dipole moment $M$ . The angle of dip $δ$ is related to latitude $λ$ as

1

\tan δ = \frac{\tan λ}{2}

2

\tan δ = 2 \tan λ

3

\tan δ - \cot λ

4

\tan δ = \tan λ

Explanation:

$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos θ, B_{θ} = \frac{μ_{0} M \sin θ}{4 π r^{3}}$
as
$θ = 90 - λ$
$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos (90 - λ)$
$= \frac{2 μ_{0} M \sin λ}{4 π r^{3}}$
$\frac{B_{r}}{B_{θ}} = \frac{B_{r}}{B_{θ}} = 2 \tan λ o r \tan δ = 2 \tan λ$
supporting img

PHXII05:MAGNETISM and MATTER

360686 The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{- 4} W b m^{- 2}$ . If the angle of dip at that place is $60^{\circ}$ , then the value of vertical component of the earth's magnetic field will be (in $W b m^{- 2}$ )

1

0.12 \times 10^{- 4}

2

0.24 \times 10^{- 4}

3

0.40 \times 10^{- 4}

4

0.622 \times 10^{- 4}

Explanation:

Vertical component of earth's magnetic field,
$B_{V} = B_{H} \tan δ$
$= 0.36 \times 10^{- 4} \times \tan 60^{0}$ $(g i v e n)$
$= 0.36 \times 10^{- 4} \times \sqrt{3} = 0.622 \times 10^{- 4} T$
$= 0.622 \times 10^{- 4} W b m^{- 2}$
So, correct option is (4)

AIIMS - 2018

PHXII05:MAGNETISM and MATTER

360683 At certain place, horizontal component is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip is

1

30^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

Here, horizontal component
$B_{H} = \frac{1}{\sqrt{3}} \times B_{V}$
$\tan δ = \frac{B_{V}}{B_{H}} = \frac{B_{V}}{\frac{1}{\sqrt{3}} B_{V}} = \sqrt{3}$
$\tan δ = \sqrt{3} \Rightarrow δ = \tan^{- 1} (\sqrt{3}) = 60^{\circ}$ .
So correct option is (4)

PHXII05:MAGNETISM and MATTER

360684 If $B_{H} = \frac{1}{\sqrt{3}} B_{V}$ , then find angle of dip. ( where, symbols have their usual meanings)

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

Magnetic dip or magnetic inclination is given by
$\tan δ = \frac{B_{V}}{B_{H}} (1)$
Where, $B_{V}$ and $B_{H}$ are vertical and horizontal components of earth's magnetic field, respectively.
Given, $B_{H} = \frac{1}{\sqrt{3}} B_{V} \Rightarrow \frac{B_{V}}{B_{H}} = \sqrt{3} (2)$
From Eqs. (1) and (2), we get
$\tan δ = \sqrt{3} \Rightarrow δ = 60^{\circ}$

PHXII05:MAGNETISM and MATTER

360685 Consider the earth as a short magnet with its centre coinciding with the centre of the earth and dipole moment $M$ . The angle of dip $δ$ is related to latitude $λ$ as

1

\tan δ = \frac{\tan λ}{2}

2

\tan δ = 2 \tan λ

3

\tan δ - \cot λ

4

\tan δ = \tan λ

Explanation:

$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos θ, B_{θ} = \frac{μ_{0} M \sin θ}{4 π r^{3}}$
as
$θ = 90 - λ$
$B_{r} = \frac{2 μ_{0} M}{4 π r^{3}} \cos (90 - λ)$
$= \frac{2 μ_{0} M \sin λ}{4 π r^{3}}$
$\frac{B_{r}}{B_{θ}} = \frac{B_{r}}{B_{θ}} = 2 \tan λ o r \tan δ = 2 \tan λ$
supporting img

PHXII05:MAGNETISM and MATTER

360686 The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{- 4} W b m^{- 2}$ . If the angle of dip at that place is $60^{\circ}$ , then the value of vertical component of the earth's magnetic field will be (in $W b m^{- 2}$ )

1

0.12 \times 10^{- 4}

2

0.24 \times 10^{- 4}

3

0.40 \times 10^{- 4}

4

0.622 \times 10^{- 4}

Explanation:

Vertical component of earth's magnetic field,
$B_{V} = B_{H} \tan δ$
$= 0.36 \times 10^{- 4} \times \tan 60^{0}$ $(g i v e n)$
$= 0.36 \times 10^{- 4} \times \sqrt{3} = 0.622 \times 10^{- 4} T$
$= 0.622 \times 10^{- 4} W b m^{- 2}$
So, correct option is (4)

AIIMS - 2018

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: