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Magnetism and Matter

154109 The total intensity of earth's magnetic field at the poles is $7$ units. Its value at the equator is

1

7 \sqrt{2}

units

2 3.5 units

3 7 units

4

\frac{7}{\sqrt{2}}

units

5 14 units

Explanation:

C Given,
Total intensity of earth's magnetic field of the poles, $B_{V}$ $= 7$ units
We know that,
$B_{V} = B \cos θ [At equator, θ = 0^{\circ}]$
$B = \frac{B_{V}}{\cos θ}$
$B = \frac{7}{1}$
$B = 7 units$

Kerala CEE 2021

Magnetism and Matter

154110 Torques $τ_{1}$ and $τ_{2}$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_{1}$ and $B_{2}$ at two different places. The ratio $B_{1} / B_{2}$ is

1

τ_{2} / τ_{1}

2

τ_{1} / τ_{2}

3

\frac{τ_{1} + τ_{2}}{τ_{1} - τ_{2}}

4

\frac{τ_{1} - τ_{2}}{τ_{1} + τ_{2}}

Explanation:

B We know that,
$τ = MB \sin θ$
When, $θ = 90^{\circ}$
So, $τ = MB$
Then, $τ_{1} = {MB}_{1}$
And, $τ_{2} = {MB}_{2}$
$∴ \frac{τ_{1}}{τ_{2}} = \frac{{MB}_{1}}{{MB}_{2}}$
$∴ \frac{B_{1}}{B_{2}} = \frac{τ_{1}}{τ_{2}}$
Magnetism and Matter

AP EAMCET-24.08.2021

Magnetism and Matter

154111 The earth's magnetic field at a certain place has a horizontal component $0.3 G$ and the total strength $0.5 G$ . The angle of dip is

1

\tan^{- 1} (\frac{3}{4})

2

\sin^{- 1} (\frac{3}{4})

3

\tan^{- 1} (\frac{4}{3})

4

\sin^{- 1} (\frac{4}{3})

Explanation:

C Given that,
Horizontal component of earth's field $(B_{H}) = 0.3 G$
Total magnetic field intensity $(B) = 0.5 G$
Now, $B = \sqrt{B_{H}^{2} + B_{V}^{2}}$
$B_{v} = \sqrt{B^{2} - B_{H}^{2}}$
Where, $B_{V} =$ Vertical component
$∴ B_{V} = \sqrt{(0.5)^{2} - (0.3)^{2}} = \sqrt{0.16} = 0.4 G$
Angle of dip, $\tan θ = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3}$
$θ = \tan^{- 1} (\frac{4}{3})$

AP EAMCET-03.09.2021

Magnetism and Matter

154112 The total intensity of the earth's magnetic field at equator is 5 units. What is its value at the poles?

1 5

2 4

3 3

4 2

Explanation:

A Given,
Total intensity of the earth's magnetic field at equator, $B = 5$ units At equator, $θ = 0^{\circ}$
Horizontal component of earths field,
$B_{H} = B \cos θ$
$B_{H} = 5 \cos 0^{\circ}$
$B_{H} = 5$ Units
At poles, $θ = 90^{\circ}$
Vertical components of earth's field,
$B_{v} = B \sin θ$
$B_{v} = 5 \sin 90^{\circ}$
$B_{v} = 5 Units$
So, the intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction,

AP EAMCET-06.09.2021

Magnetism and Matter

154113 A magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of earth's magnetic field is $20 μ T$ . If the neutral point is found at a distance of $20 cm$ in the plane bisecting the dipole then the magnetic moment of the dipole is:
(Assume $μ_{0} = 4 π \times 10^{- 7} S$ .I units)

1

1.2 {Am}^{2}

2

2.2 {Am}^{2}

3

1.4 {Am}^{2}

4

1.6 {Am}^{2}

Explanation:

D Given, $B = 20 μ T = 20 \times 10^{- 6} T$
$d = 20 cm = 20 \times 10^{- 2} m, M = ?$
We know that, $B = \frac{μ_{0} M}{4 π d^{3}}$
$20 \times 10^{- 6} = \frac{10^{- 7} \times M}{{(20 \times 10^{- 2})}^{3}}$
$M \times 10^{- 7} = 20 \times 20 \times 20 \times 20 \times 10^{- 12}$
$M = \frac{16 \times 10^{4} \times 10^{- 12}}{10^{- 7}}$
$M = 16 \times 10^{- 8} \times 10^{7}$
$M = 1.6 {Am}^{2}$

TS EAMCET 05.08.2021

Magnetism and Matter

154109 The total intensity of earth's magnetic field at the poles is $7$ units. Its value at the equator is

1

7 \sqrt{2}

units

2 3.5 units

3 7 units

4

\frac{7}{\sqrt{2}}

units

5 14 units

Explanation:

C Given,
Total intensity of earth's magnetic field of the poles, $B_{V}$ $= 7$ units
We know that,
$B_{V} = B \cos θ [At equator, θ = 0^{\circ}]$
$B = \frac{B_{V}}{\cos θ}$
$B = \frac{7}{1}$
$B = 7 units$

Kerala CEE 2021

Magnetism and Matter

154110 Torques $τ_{1}$ and $τ_{2}$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_{1}$ and $B_{2}$ at two different places. The ratio $B_{1} / B_{2}$ is

1

τ_{2} / τ_{1}

2

τ_{1} / τ_{2}

3

\frac{τ_{1} + τ_{2}}{τ_{1} - τ_{2}}

4

\frac{τ_{1} - τ_{2}}{τ_{1} + τ_{2}}

Explanation:

B We know that,
$τ = MB \sin θ$
When, $θ = 90^{\circ}$
So, $τ = MB$
Then, $τ_{1} = {MB}_{1}$
And, $τ_{2} = {MB}_{2}$
$∴ \frac{τ_{1}}{τ_{2}} = \frac{{MB}_{1}}{{MB}_{2}}$
$∴ \frac{B_{1}}{B_{2}} = \frac{τ_{1}}{τ_{2}}$
Magnetism and Matter

AP EAMCET-24.08.2021

Magnetism and Matter

154111 The earth's magnetic field at a certain place has a horizontal component $0.3 G$ and the total strength $0.5 G$ . The angle of dip is

1

\tan^{- 1} (\frac{3}{4})

2

\sin^{- 1} (\frac{3}{4})

3

\tan^{- 1} (\frac{4}{3})

4

\sin^{- 1} (\frac{4}{3})

Explanation:

C Given that,
Horizontal component of earth's field $(B_{H}) = 0.3 G$
Total magnetic field intensity $(B) = 0.5 G$
Now, $B = \sqrt{B_{H}^{2} + B_{V}^{2}}$
$B_{v} = \sqrt{B^{2} - B_{H}^{2}}$
Where, $B_{V} =$ Vertical component
$∴ B_{V} = \sqrt{(0.5)^{2} - (0.3)^{2}} = \sqrt{0.16} = 0.4 G$
Angle of dip, $\tan θ = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3}$
$θ = \tan^{- 1} (\frac{4}{3})$

AP EAMCET-03.09.2021

Magnetism and Matter

154112 The total intensity of the earth's magnetic field at equator is 5 units. What is its value at the poles?

1 5

2 4

3 3

4 2

Explanation:

A Given,
Total intensity of the earth's magnetic field at equator, $B = 5$ units At equator, $θ = 0^{\circ}$
Horizontal component of earths field,
$B_{H} = B \cos θ$
$B_{H} = 5 \cos 0^{\circ}$
$B_{H} = 5$ Units
At poles, $θ = 90^{\circ}$
Vertical components of earth's field,
$B_{v} = B \sin θ$
$B_{v} = 5 \sin 90^{\circ}$
$B_{v} = 5 Units$
So, the intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction,

AP EAMCET-06.09.2021

Magnetism and Matter

154113 A magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of earth's magnetic field is $20 μ T$ . If the neutral point is found at a distance of $20 cm$ in the plane bisecting the dipole then the magnetic moment of the dipole is:
(Assume $μ_{0} = 4 π \times 10^{- 7} S$ .I units)

1

1.2 {Am}^{2}

2

2.2 {Am}^{2}

3

1.4 {Am}^{2}

4

1.6 {Am}^{2}

Explanation:

D Given, $B = 20 μ T = 20 \times 10^{- 6} T$
$d = 20 cm = 20 \times 10^{- 2} m, M = ?$
We know that, $B = \frac{μ_{0} M}{4 π d^{3}}$
$20 \times 10^{- 6} = \frac{10^{- 7} \times M}{{(20 \times 10^{- 2})}^{3}}$
$M \times 10^{- 7} = 20 \times 20 \times 20 \times 20 \times 10^{- 12}$
$M = \frac{16 \times 10^{4} \times 10^{- 12}}{10^{- 7}}$
$M = 16 \times 10^{- 8} \times 10^{7}$
$M = 1.6 {Am}^{2}$

TS EAMCET 05.08.2021

Magnetism and Matter

154109 The total intensity of earth's magnetic field at the poles is $7$ units. Its value at the equator is

1

7 \sqrt{2}

units

2 3.5 units

3 7 units

4

\frac{7}{\sqrt{2}}

units

5 14 units

Explanation:

C Given,
Total intensity of earth's magnetic field of the poles, $B_{V}$ $= 7$ units
We know that,
$B_{V} = B \cos θ [At equator, θ = 0^{\circ}]$
$B = \frac{B_{V}}{\cos θ}$
$B = \frac{7}{1}$
$B = 7 units$

Kerala CEE 2021

Magnetism and Matter

154110 Torques $τ_{1}$ and $τ_{2}$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_{1}$ and $B_{2}$ at two different places. The ratio $B_{1} / B_{2}$ is

1

τ_{2} / τ_{1}

2

τ_{1} / τ_{2}

3

\frac{τ_{1} + τ_{2}}{τ_{1} - τ_{2}}

4

\frac{τ_{1} - τ_{2}}{τ_{1} + τ_{2}}

Explanation:

B We know that,
$τ = MB \sin θ$
When, $θ = 90^{\circ}$
So, $τ = MB$
Then, $τ_{1} = {MB}_{1}$
And, $τ_{2} = {MB}_{2}$
$∴ \frac{τ_{1}}{τ_{2}} = \frac{{MB}_{1}}{{MB}_{2}}$
$∴ \frac{B_{1}}{B_{2}} = \frac{τ_{1}}{τ_{2}}$
Magnetism and Matter

AP EAMCET-24.08.2021

Magnetism and Matter

154111 The earth's magnetic field at a certain place has a horizontal component $0.3 G$ and the total strength $0.5 G$ . The angle of dip is

1

\tan^{- 1} (\frac{3}{4})

2

\sin^{- 1} (\frac{3}{4})

3

\tan^{- 1} (\frac{4}{3})

4

\sin^{- 1} (\frac{4}{3})

Explanation:

C Given that,
Horizontal component of earth's field $(B_{H}) = 0.3 G$
Total magnetic field intensity $(B) = 0.5 G$
Now, $B = \sqrt{B_{H}^{2} + B_{V}^{2}}$
$B_{v} = \sqrt{B^{2} - B_{H}^{2}}$
Where, $B_{V} =$ Vertical component
$∴ B_{V} = \sqrt{(0.5)^{2} - (0.3)^{2}} = \sqrt{0.16} = 0.4 G$
Angle of dip, $\tan θ = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3}$
$θ = \tan^{- 1} (\frac{4}{3})$

AP EAMCET-03.09.2021

Magnetism and Matter

154112 The total intensity of the earth's magnetic field at equator is 5 units. What is its value at the poles?

1 5

2 4

3 3

4 2

Explanation:

A Given,
Total intensity of the earth's magnetic field at equator, $B = 5$ units At equator, $θ = 0^{\circ}$
Horizontal component of earths field,
$B_{H} = B \cos θ$
$B_{H} = 5 \cos 0^{\circ}$
$B_{H} = 5$ Units
At poles, $θ = 90^{\circ}$
Vertical components of earth's field,
$B_{v} = B \sin θ$
$B_{v} = 5 \sin 90^{\circ}$
$B_{v} = 5 Units$
So, the intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction,

AP EAMCET-06.09.2021

Magnetism and Matter

154113 A magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of earth's magnetic field is $20 μ T$ . If the neutral point is found at a distance of $20 cm$ in the plane bisecting the dipole then the magnetic moment of the dipole is:
(Assume $μ_{0} = 4 π \times 10^{- 7} S$ .I units)

1

1.2 {Am}^{2}

2

2.2 {Am}^{2}

3

1.4 {Am}^{2}

4

1.6 {Am}^{2}

Explanation:

D Given, $B = 20 μ T = 20 \times 10^{- 6} T$
$d = 20 cm = 20 \times 10^{- 2} m, M = ?$
We know that, $B = \frac{μ_{0} M}{4 π d^{3}}$
$20 \times 10^{- 6} = \frac{10^{- 7} \times M}{{(20 \times 10^{- 2})}^{3}}$
$M \times 10^{- 7} = 20 \times 20 \times 20 \times 20 \times 10^{- 12}$
$M = \frac{16 \times 10^{4} \times 10^{- 12}}{10^{- 7}}$
$M = 16 \times 10^{- 8} \times 10^{7}$
$M = 1.6 {Am}^{2}$

TS EAMCET 05.08.2021

Magnetism and Matter

154109 The total intensity of earth's magnetic field at the poles is $7$ units. Its value at the equator is

1

7 \sqrt{2}

units

2 3.5 units

3 7 units

4

\frac{7}{\sqrt{2}}

units

5 14 units

Explanation:

C Given,
Total intensity of earth's magnetic field of the poles, $B_{V}$ $= 7$ units
We know that,
$B_{V} = B \cos θ [At equator, θ = 0^{\circ}]$
$B = \frac{B_{V}}{\cos θ}$
$B = \frac{7}{1}$
$B = 7 units$

Kerala CEE 2021

Magnetism and Matter

154110 Torques $τ_{1}$ and $τ_{2}$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_{1}$ and $B_{2}$ at two different places. The ratio $B_{1} / B_{2}$ is

1

τ_{2} / τ_{1}

2

τ_{1} / τ_{2}

3

\frac{τ_{1} + τ_{2}}{τ_{1} - τ_{2}}

4

\frac{τ_{1} - τ_{2}}{τ_{1} + τ_{2}}

Explanation:

B We know that,
$τ = MB \sin θ$
When, $θ = 90^{\circ}$
So, $τ = MB$
Then, $τ_{1} = {MB}_{1}$
And, $τ_{2} = {MB}_{2}$
$∴ \frac{τ_{1}}{τ_{2}} = \frac{{MB}_{1}}{{MB}_{2}}$
$∴ \frac{B_{1}}{B_{2}} = \frac{τ_{1}}{τ_{2}}$
Magnetism and Matter

AP EAMCET-24.08.2021

Magnetism and Matter

154111 The earth's magnetic field at a certain place has a horizontal component $0.3 G$ and the total strength $0.5 G$ . The angle of dip is

1

\tan^{- 1} (\frac{3}{4})

2

\sin^{- 1} (\frac{3}{4})

3

\tan^{- 1} (\frac{4}{3})

4

\sin^{- 1} (\frac{4}{3})

Explanation:

C Given that,
Horizontal component of earth's field $(B_{H}) = 0.3 G$
Total magnetic field intensity $(B) = 0.5 G$
Now, $B = \sqrt{B_{H}^{2} + B_{V}^{2}}$
$B_{v} = \sqrt{B^{2} - B_{H}^{2}}$
Where, $B_{V} =$ Vertical component
$∴ B_{V} = \sqrt{(0.5)^{2} - (0.3)^{2}} = \sqrt{0.16} = 0.4 G$
Angle of dip, $\tan θ = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3}$
$θ = \tan^{- 1} (\frac{4}{3})$

AP EAMCET-03.09.2021

Magnetism and Matter

154112 The total intensity of the earth's magnetic field at equator is 5 units. What is its value at the poles?

1 5

2 4

3 3

4 2

Explanation:

A Given,
Total intensity of the earth's magnetic field at equator, $B = 5$ units At equator, $θ = 0^{\circ}$
Horizontal component of earths field,
$B_{H} = B \cos θ$
$B_{H} = 5 \cos 0^{\circ}$
$B_{H} = 5$ Units
At poles, $θ = 90^{\circ}$
Vertical components of earth's field,
$B_{v} = B \sin θ$
$B_{v} = 5 \sin 90^{\circ}$
$B_{v} = 5 Units$
So, the intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction,

AP EAMCET-06.09.2021

Magnetism and Matter

154113 A magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of earth's magnetic field is $20 μ T$ . If the neutral point is found at a distance of $20 cm$ in the plane bisecting the dipole then the magnetic moment of the dipole is:
(Assume $μ_{0} = 4 π \times 10^{- 7} S$ .I units)

1

1.2 {Am}^{2}

2

2.2 {Am}^{2}

3

1.4 {Am}^{2}

4

1.6 {Am}^{2}

Explanation:

D Given, $B = 20 μ T = 20 \times 10^{- 6} T$
$d = 20 cm = 20 \times 10^{- 2} m, M = ?$
We know that, $B = \frac{μ_{0} M}{4 π d^{3}}$
$20 \times 10^{- 6} = \frac{10^{- 7} \times M}{{(20 \times 10^{- 2})}^{3}}$
$M \times 10^{- 7} = 20 \times 20 \times 20 \times 20 \times 10^{- 12}$
$M = \frac{16 \times 10^{4} \times 10^{- 12}}{10^{- 7}}$
$M = 16 \times 10^{- 8} \times 10^{7}$
$M = 1.6 {Am}^{2}$

TS EAMCET 05.08.2021

Magnetism and Matter

154109 The total intensity of earth's magnetic field at the poles is $7$ units. Its value at the equator is

1

7 \sqrt{2}

units

2 3.5 units

3 7 units

4

\frac{7}{\sqrt{2}}

units

5 14 units

Explanation:

C Given,
Total intensity of earth's magnetic field of the poles, $B_{V}$ $= 7$ units
We know that,
$B_{V} = B \cos θ [At equator, θ = 0^{\circ}]$
$B = \frac{B_{V}}{\cos θ}$
$B = \frac{7}{1}$
$B = 7 units$

Kerala CEE 2021

Magnetism and Matter

154110 Torques $τ_{1}$ and $τ_{2}$ are required for a magnetic needle to remain perpendicular to the magnetic field $B_{1}$ and $B_{2}$ at two different places. The ratio $B_{1} / B_{2}$ is

1

τ_{2} / τ_{1}

2

τ_{1} / τ_{2}

3

\frac{τ_{1} + τ_{2}}{τ_{1} - τ_{2}}

4

\frac{τ_{1} - τ_{2}}{τ_{1} + τ_{2}}

Explanation:

B We know that,
$τ = MB \sin θ$
When, $θ = 90^{\circ}$
So, $τ = MB$
Then, $τ_{1} = {MB}_{1}$
And, $τ_{2} = {MB}_{2}$
$∴ \frac{τ_{1}}{τ_{2}} = \frac{{MB}_{1}}{{MB}_{2}}$
$∴ \frac{B_{1}}{B_{2}} = \frac{τ_{1}}{τ_{2}}$
Magnetism and Matter

AP EAMCET-24.08.2021

Magnetism and Matter

154111 The earth's magnetic field at a certain place has a horizontal component $0.3 G$ and the total strength $0.5 G$ . The angle of dip is

1

\tan^{- 1} (\frac{3}{4})

2

\sin^{- 1} (\frac{3}{4})

3

\tan^{- 1} (\frac{4}{3})

4

\sin^{- 1} (\frac{4}{3})

Explanation:

C Given that,
Horizontal component of earth's field $(B_{H}) = 0.3 G$
Total magnetic field intensity $(B) = 0.5 G$
Now, $B = \sqrt{B_{H}^{2} + B_{V}^{2}}$
$B_{v} = \sqrt{B^{2} - B_{H}^{2}}$
Where, $B_{V} =$ Vertical component
$∴ B_{V} = \sqrt{(0.5)^{2} - (0.3)^{2}} = \sqrt{0.16} = 0.4 G$
Angle of dip, $\tan θ = \frac{B_{V}}{B_{H}} = \frac{0.4}{0.3}$
$θ = \tan^{- 1} (\frac{4}{3})$

AP EAMCET-03.09.2021

Magnetism and Matter

154112 The total intensity of the earth's magnetic field at equator is 5 units. What is its value at the poles?

1 5

2 4

3 3

4 2

Explanation:

A Given,
Total intensity of the earth's magnetic field at equator, $B = 5$ units At equator, $θ = 0^{\circ}$
Horizontal component of earths field,
$B_{H} = B \cos θ$
$B_{H} = 5 \cos 0^{\circ}$
$B_{H} = 5$ Units
At poles, $θ = 90^{\circ}$
Vertical components of earth's field,
$B_{v} = B \sin θ$
$B_{v} = 5 \sin 90^{\circ}$
$B_{v} = 5 Units$
So, the intensity of earth's magnetic field at poles is also 5 units, but in the vertical direction,

AP EAMCET-06.09.2021

Magnetism and Matter

154113 A magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of earth's magnetic field is $20 μ T$ . If the neutral point is found at a distance of $20 cm$ in the plane bisecting the dipole then the magnetic moment of the dipole is:
(Assume $μ_{0} = 4 π \times 10^{- 7} S$ .I units)

1

1.2 {Am}^{2}

2

2.2 {Am}^{2}

3

1.4 {Am}^{2}

4

1.6 {Am}^{2}

Explanation:

D Given, $B = 20 μ T = 20 \times 10^{- 6} T$
$d = 20 cm = 20 \times 10^{- 2} m, M = ?$
We know that, $B = \frac{μ_{0} M}{4 π d^{3}}$
$20 \times 10^{- 6} = \frac{10^{- 7} \times M}{{(20 \times 10^{- 2})}^{3}}$
$M \times 10^{- 7} = 20 \times 20 \times 20 \times 20 \times 10^{- 12}$
$M = \frac{16 \times 10^{4} \times 10^{- 12}}{10^{- 7}}$
$M = 16 \times 10^{- 8} \times 10^{7}$
$M = 1.6 {Am}^{2}$

TS EAMCET 05.08.2021

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