QBManager

Magnetism and Matter

154116 The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is

1

45^{\circ}

2

30^{\circ}

3

0^{\circ}

4

60^{\circ}

5

90^{\circ}

Explanation:

A We know that,
Vertical component of earth's magnetic field is , $B_{V} = B \sin θ$
Horizontal component of earth's magnetic field is, $B_{H} = B \cos θ$
According to question,
$B_{H} = B_{V}$
$B \cos θ = B \sin θ$
$\frac{B \sin θ}{B \cos θ} = 1$
$\tan θ = 1$
$\tan θ = \tan 45^{\circ}$
$θ = 45^{\circ}$

Kerala CEE- 2021

Magnetism and Matter

154117 The angle of dip at a place is $30^{\circ}$ . If horizontal component of earth's magnetic field is $H$ , the total field intensity is

1

\frac{H}{\sqrt{2}}

2

\frac{2 H}{\sqrt{3}}

3

H \sqrt{2}

4

H \sqrt{3}

Explanation:

B Given,
Angle of dip $(θ) = 30^{\circ}$
Horizontal component of earth's magnetic field,
$B_{H} = H$
We know that, $B_{H} = B_{e} \cos θ$
$B_{e} = \frac{B_{H}}{Cos 30^{\circ}}$
$B_{e} = \frac{2 H}{\sqrt{3}}$

UPSEE 2020

Magnetism and Matter

154120 In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$ . What is the magnetic field of the earth at this location?

1

0.4 G

2

0.2 G

3

\frac{2 \sqrt{3}}{5} G

4

\frac{\sqrt{2}}{5} G

Explanation:

A Given that,
$B_{H} = \frac{\sqrt{3}}{5} G, dip angle (δ) = 30^{\circ}$
We know that,
$B_{H} = B \cos δ$
$\cos δ = \frac{B_{H}}{B}$
Then, $\cos 30^{\circ} = \frac{B_{H}}{B}$
$\frac{\sqrt{3}}{2} = \frac{\frac{\sqrt{3}}{5} G}{B}$
$B = \frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}} = \frac{2}{5}$
So, $B = 0.4 G$

TS EAMCET 29.09.2020

Magnetism and Matter

154121 A coil of 50 turns and $10 cm$ diameter has a resistance of $10 Ω$ . What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H} = 0.314 G$ at the centre of the coil?

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
$B_{H} = 0.314 G$
$B_{H} = 0.314 \times 10^{- 4} T$
The current in coil be 'I' then the magnetic field at the centre of the coil is -
$B = \frac{μ_{0} In}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times I \times 50}{2 \times 5 \times 10^{- 2}}$
$B = 4 π \times 10^{- 5} \times 5 \times I$
$I = \frac{B}{4 π \times 10^{- 5} \times 5}$
$I = \frac{0.314 \times 10^{- 4}}{4 π \times 10^{- 5} \times 5}$
$I = \frac{1}{20} \Rightarrow 0.05 A$
$∴$ Potential difference $(V) = IR$
$= 0.05 \times 10 = 0.5 V$

Assam CEE-2020

Magnetism and Matter

154116 The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is

1

45^{\circ}

2

30^{\circ}

3

0^{\circ}

4

60^{\circ}

5

90^{\circ}

Explanation:

A We know that,
Vertical component of earth's magnetic field is , $B_{V} = B \sin θ$
Horizontal component of earth's magnetic field is, $B_{H} = B \cos θ$
According to question,
$B_{H} = B_{V}$
$B \cos θ = B \sin θ$
$\frac{B \sin θ}{B \cos θ} = 1$
$\tan θ = 1$
$\tan θ = \tan 45^{\circ}$
$θ = 45^{\circ}$

Kerala CEE- 2021

Magnetism and Matter

154117 The angle of dip at a place is $30^{\circ}$ . If horizontal component of earth's magnetic field is $H$ , the total field intensity is

1

\frac{H}{\sqrt{2}}

2

\frac{2 H}{\sqrt{3}}

3

H \sqrt{2}

4

H \sqrt{3}

Explanation:

B Given,
Angle of dip $(θ) = 30^{\circ}$
Horizontal component of earth's magnetic field,
$B_{H} = H$
We know that, $B_{H} = B_{e} \cos θ$
$B_{e} = \frac{B_{H}}{Cos 30^{\circ}}$
$B_{e} = \frac{2 H}{\sqrt{3}}$

UPSEE 2020

Magnetism and Matter

154120 In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$ . What is the magnetic field of the earth at this location?

1

0.4 G

2

0.2 G

3

\frac{2 \sqrt{3}}{5} G

4

\frac{\sqrt{2}}{5} G

Explanation:

A Given that,
$B_{H} = \frac{\sqrt{3}}{5} G, dip angle (δ) = 30^{\circ}$
We know that,
$B_{H} = B \cos δ$
$\cos δ = \frac{B_{H}}{B}$
Then, $\cos 30^{\circ} = \frac{B_{H}}{B}$
$\frac{\sqrt{3}}{2} = \frac{\frac{\sqrt{3}}{5} G}{B}$
$B = \frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}} = \frac{2}{5}$
So, $B = 0.4 G$

TS EAMCET 29.09.2020

Magnetism and Matter

154121 A coil of 50 turns and $10 cm$ diameter has a resistance of $10 Ω$ . What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H} = 0.314 G$ at the centre of the coil?

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
$B_{H} = 0.314 G$
$B_{H} = 0.314 \times 10^{- 4} T$
The current in coil be 'I' then the magnetic field at the centre of the coil is -
$B = \frac{μ_{0} In}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times I \times 50}{2 \times 5 \times 10^{- 2}}$
$B = 4 π \times 10^{- 5} \times 5 \times I$
$I = \frac{B}{4 π \times 10^{- 5} \times 5}$
$I = \frac{0.314 \times 10^{- 4}}{4 π \times 10^{- 5} \times 5}$
$I = \frac{1}{20} \Rightarrow 0.05 A$
$∴$ Potential difference $(V) = IR$
$= 0.05 \times 10 = 0.5 V$

Assam CEE-2020

Magnetism and Matter

154116 The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is

1

45^{\circ}

2

30^{\circ}

3

0^{\circ}

4

60^{\circ}

5

90^{\circ}

Explanation:

A We know that,
Vertical component of earth's magnetic field is , $B_{V} = B \sin θ$
Horizontal component of earth's magnetic field is, $B_{H} = B \cos θ$
According to question,
$B_{H} = B_{V}$
$B \cos θ = B \sin θ$
$\frac{B \sin θ}{B \cos θ} = 1$
$\tan θ = 1$
$\tan θ = \tan 45^{\circ}$
$θ = 45^{\circ}$

Kerala CEE- 2021

Magnetism and Matter

154117 The angle of dip at a place is $30^{\circ}$ . If horizontal component of earth's magnetic field is $H$ , the total field intensity is

1

\frac{H}{\sqrt{2}}

2

\frac{2 H}{\sqrt{3}}

3

H \sqrt{2}

4

H \sqrt{3}

Explanation:

B Given,
Angle of dip $(θ) = 30^{\circ}$
Horizontal component of earth's magnetic field,
$B_{H} = H$
We know that, $B_{H} = B_{e} \cos θ$
$B_{e} = \frac{B_{H}}{Cos 30^{\circ}}$
$B_{e} = \frac{2 H}{\sqrt{3}}$

UPSEE 2020

Magnetism and Matter

154120 In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$ . What is the magnetic field of the earth at this location?

1

0.4 G

2

0.2 G

3

\frac{2 \sqrt{3}}{5} G

4

\frac{\sqrt{2}}{5} G

Explanation:

A Given that,
$B_{H} = \frac{\sqrt{3}}{5} G, dip angle (δ) = 30^{\circ}$
We know that,
$B_{H} = B \cos δ$
$\cos δ = \frac{B_{H}}{B}$
Then, $\cos 30^{\circ} = \frac{B_{H}}{B}$
$\frac{\sqrt{3}}{2} = \frac{\frac{\sqrt{3}}{5} G}{B}$
$B = \frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}} = \frac{2}{5}$
So, $B = 0.4 G$

TS EAMCET 29.09.2020

Magnetism and Matter

154121 A coil of 50 turns and $10 cm$ diameter has a resistance of $10 Ω$ . What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H} = 0.314 G$ at the centre of the coil?

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
$B_{H} = 0.314 G$
$B_{H} = 0.314 \times 10^{- 4} T$
The current in coil be 'I' then the magnetic field at the centre of the coil is -
$B = \frac{μ_{0} In}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times I \times 50}{2 \times 5 \times 10^{- 2}}$
$B = 4 π \times 10^{- 5} \times 5 \times I$
$I = \frac{B}{4 π \times 10^{- 5} \times 5}$
$I = \frac{0.314 \times 10^{- 4}}{4 π \times 10^{- 5} \times 5}$
$I = \frac{1}{20} \Rightarrow 0.05 A$
$∴$ Potential difference $(V) = IR$
$= 0.05 \times 10 = 0.5 V$

Assam CEE-2020

Magnetism and Matter

154116 The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is

1

45^{\circ}

2

30^{\circ}

3

0^{\circ}

4

60^{\circ}

5

90^{\circ}

Explanation:

A We know that,
Vertical component of earth's magnetic field is , $B_{V} = B \sin θ$
Horizontal component of earth's magnetic field is, $B_{H} = B \cos θ$
According to question,
$B_{H} = B_{V}$
$B \cos θ = B \sin θ$
$\frac{B \sin θ}{B \cos θ} = 1$
$\tan θ = 1$
$\tan θ = \tan 45^{\circ}$
$θ = 45^{\circ}$

Kerala CEE- 2021

Magnetism and Matter

154117 The angle of dip at a place is $30^{\circ}$ . If horizontal component of earth's magnetic field is $H$ , the total field intensity is

1

\frac{H}{\sqrt{2}}

2

\frac{2 H}{\sqrt{3}}

3

H \sqrt{2}

4

H \sqrt{3}

Explanation:

B Given,
Angle of dip $(θ) = 30^{\circ}$
Horizontal component of earth's magnetic field,
$B_{H} = H$
We know that, $B_{H} = B_{e} \cos θ$
$B_{e} = \frac{B_{H}}{Cos 30^{\circ}}$
$B_{e} = \frac{2 H}{\sqrt{3}}$

UPSEE 2020

Magnetism and Matter

154120 In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$ . What is the magnetic field of the earth at this location?

1

0.4 G

2

0.2 G

3

\frac{2 \sqrt{3}}{5} G

4

\frac{\sqrt{2}}{5} G

Explanation:

A Given that,
$B_{H} = \frac{\sqrt{3}}{5} G, dip angle (δ) = 30^{\circ}$
We know that,
$B_{H} = B \cos δ$
$\cos δ = \frac{B_{H}}{B}$
Then, $\cos 30^{\circ} = \frac{B_{H}}{B}$
$\frac{\sqrt{3}}{2} = \frac{\frac{\sqrt{3}}{5} G}{B}$
$B = \frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}} = \frac{2}{5}$
So, $B = 0.4 G$

TS EAMCET 29.09.2020

Magnetism and Matter

154121 A coil of 50 turns and $10 cm$ diameter has a resistance of $10 Ω$ . What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H} = 0.314 G$ at the centre of the coil?

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
$B_{H} = 0.314 G$
$B_{H} = 0.314 \times 10^{- 4} T$
The current in coil be 'I' then the magnetic field at the centre of the coil is -
$B = \frac{μ_{0} In}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times I \times 50}{2 \times 5 \times 10^{- 2}}$
$B = 4 π \times 10^{- 5} \times 5 \times I$
$I = \frac{B}{4 π \times 10^{- 5} \times 5}$
$I = \frac{0.314 \times 10^{- 4}}{4 π \times 10^{- 5} \times 5}$
$I = \frac{1}{20} \Rightarrow 0.05 A$
$∴$ Potential difference $(V) = IR$
$= 0.05 \times 10 = 0.5 V$

Assam CEE-2020

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: