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Moving Charges & Magnetism

153920 Magnetic moment due to the motion of the electron in $n^{th}$ energy state of hydrogen atom is proportional to .............

1

n^{- 2}

2

n

3

n^{2}

4

n^{3}

Explanation:

B The magnetic moment due to the motion of electron in $n^{th}$ energy state of hydrogen atom is proportional to $n$ .
We know that,
$\frac{Magnetic moment}{Angular momentum} = \frac{e}{2 m}$
$\frac{M}{L} = \frac{e}{2 m}$
$M = \frac{L e}{2 m}$
Hence, $M \propto L$
Since, $L = \frac{nh}{2 π}$
So, $M \propto L \propto n$

AP EAMCET (21.04.2019) Shift-I

Moving Charges & Magnetism

153921 A coil in the shape of an equilateral triangle of side $2 cm$ is suspended from a vertex such that its hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{- 3}$ T. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{- 5} Nm$ , the current to be passed through the coil is

1

0.5 A

2

1 A

3

2 A

4

4 A

Explanation:

C Given :
$a = 2 \times 10^{- 2} m$
$A = \frac{\sqrt{3}}{4} \times a^{2}$
$A = \sqrt{3} \times 10^{- 4} m^{2}$
$∴ B = 100 \times 10^{- 3} T = 0.1 T$
$τ = 2 \sqrt{3} \times 10^{- 5} m$
We know that,
$\vec{τ} = \vec{M} \times \vec{B}$
$2 \sqrt{3} \times 10^{- 5} = I \times \sqrt{3} \times 10^{- 4} \times 10^{- 1} (M = I \times A)$
$I = 2 A$

AP EAMCET (20.04.2019) Shift-II

Moving Charges & Magnetism

153922 The resultant magnetic moment of three magnetic dipoles, each of the magnetic moment $M$ shown in the arrangement is

1

\sqrt{2} M

2

(\sqrt{2} + 1) M

3

(\sqrt{2} - 1) M

4

M

Explanation:

B Three magnetic moments of magnetic dipole, are shown in the figure;

$M_{1} = \sqrt{M_{A}^{2} + M_{B}^{2} + 2 M_{A} M_{B} \cos 90^{\circ}}$
$M_{1} = \sqrt{M^{2} + M^{2} + 0}$
$M_{1} = M \sqrt{2}$
The resultant of magnetic moment,
$M_{R} = M_{1} + M_{C}$
direction)
$(∵ M_{1} & M_{C} are in the same$
$M_{R} = M \sqrt{2} + M$
$M_{R} = (\sqrt{2} + 1) M$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153923 Magnetic field at the centre of a circular loop of area $A$ is $B$ . then the magnetic moment of the loop is ( $μ_{0}$ -permeability of the free space)

1

\frac{{BA}^{2}}{μ_{0} π}

2

\frac{BA \sqrt{A}}{μ_{0}}

3

\frac{BA \sqrt{A}}{μ_{0} π}

4

\frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}

Explanation:

D If $r$ is the radius of the circular loop then the area of the loop $(A) = π r^{2}$
$∴ r = \sqrt{\frac{A}{π}}$
The magnetic field at the centre of the loop,
$B = \frac{μ_{o}}{4 π} \times \frac{2 π I}{r} = \frac{μ_{o} I}{2 r}$
$I = \frac{2 Br}{μ_{o}}$
Magnetic moment,
$M = I \cdot A$
$M = \frac{2 Br}{μ_{o}} \cdot {(\frac{A}{π})}^{1 / 2}$
Putting the values of I and $r$ we get -
$M = \frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153920 Magnetic moment due to the motion of the electron in $n^{th}$ energy state of hydrogen atom is proportional to .............

1

n^{- 2}

2

n

3

n^{2}

4

n^{3}

Explanation:

B The magnetic moment due to the motion of electron in $n^{th}$ energy state of hydrogen atom is proportional to $n$ .
We know that,
$\frac{Magnetic moment}{Angular momentum} = \frac{e}{2 m}$
$\frac{M}{L} = \frac{e}{2 m}$
$M = \frac{L e}{2 m}$
Hence, $M \propto L$
Since, $L = \frac{nh}{2 π}$
So, $M \propto L \propto n$

AP EAMCET (21.04.2019) Shift-I

Moving Charges & Magnetism

153921 A coil in the shape of an equilateral triangle of side $2 cm$ is suspended from a vertex such that its hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{- 3}$ T. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{- 5} Nm$ , the current to be passed through the coil is

1

0.5 A

2

1 A

3

2 A

4

4 A

Explanation:

C Given :
$a = 2 \times 10^{- 2} m$
$A = \frac{\sqrt{3}}{4} \times a^{2}$
$A = \sqrt{3} \times 10^{- 4} m^{2}$
$∴ B = 100 \times 10^{- 3} T = 0.1 T$
$τ = 2 \sqrt{3} \times 10^{- 5} m$
We know that,
$\vec{τ} = \vec{M} \times \vec{B}$
$2 \sqrt{3} \times 10^{- 5} = I \times \sqrt{3} \times 10^{- 4} \times 10^{- 1} (M = I \times A)$
$I = 2 A$

AP EAMCET (20.04.2019) Shift-II

Moving Charges & Magnetism

153922 The resultant magnetic moment of three magnetic dipoles, each of the magnetic moment $M$ shown in the arrangement is

1

\sqrt{2} M

2

(\sqrt{2} + 1) M

3

(\sqrt{2} - 1) M

4

M

Explanation:

B Three magnetic moments of magnetic dipole, are shown in the figure;

$M_{1} = \sqrt{M_{A}^{2} + M_{B}^{2} + 2 M_{A} M_{B} \cos 90^{\circ}}$
$M_{1} = \sqrt{M^{2} + M^{2} + 0}$
$M_{1} = M \sqrt{2}$
The resultant of magnetic moment,
$M_{R} = M_{1} + M_{C}$
direction)
$(∵ M_{1} & M_{C} are in the same$
$M_{R} = M \sqrt{2} + M$
$M_{R} = (\sqrt{2} + 1) M$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153923 Magnetic field at the centre of a circular loop of area $A$ is $B$ . then the magnetic moment of the loop is ( $μ_{0}$ -permeability of the free space)

1

\frac{{BA}^{2}}{μ_{0} π}

2

\frac{BA \sqrt{A}}{μ_{0}}

3

\frac{BA \sqrt{A}}{μ_{0} π}

4

\frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}

Explanation:

D If $r$ is the radius of the circular loop then the area of the loop $(A) = π r^{2}$
$∴ r = \sqrt{\frac{A}{π}}$
The magnetic field at the centre of the loop,
$B = \frac{μ_{o}}{4 π} \times \frac{2 π I}{r} = \frac{μ_{o} I}{2 r}$
$I = \frac{2 Br}{μ_{o}}$
Magnetic moment,
$M = I \cdot A$
$M = \frac{2 Br}{μ_{o}} \cdot {(\frac{A}{π})}^{1 / 2}$
Putting the values of I and $r$ we get -
$M = \frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153920 Magnetic moment due to the motion of the electron in $n^{th}$ energy state of hydrogen atom is proportional to .............

1

n^{- 2}

2

n

3

n^{2}

4

n^{3}

Explanation:

B The magnetic moment due to the motion of electron in $n^{th}$ energy state of hydrogen atom is proportional to $n$ .
We know that,
$\frac{Magnetic moment}{Angular momentum} = \frac{e}{2 m}$
$\frac{M}{L} = \frac{e}{2 m}$
$M = \frac{L e}{2 m}$
Hence, $M \propto L$
Since, $L = \frac{nh}{2 π}$
So, $M \propto L \propto n$

AP EAMCET (21.04.2019) Shift-I

Moving Charges & Magnetism

153921 A coil in the shape of an equilateral triangle of side $2 cm$ is suspended from a vertex such that its hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{- 3}$ T. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{- 5} Nm$ , the current to be passed through the coil is

1

0.5 A

2

1 A

3

2 A

4

4 A

Explanation:

C Given :
$a = 2 \times 10^{- 2} m$
$A = \frac{\sqrt{3}}{4} \times a^{2}$
$A = \sqrt{3} \times 10^{- 4} m^{2}$
$∴ B = 100 \times 10^{- 3} T = 0.1 T$
$τ = 2 \sqrt{3} \times 10^{- 5} m$
We know that,
$\vec{τ} = \vec{M} \times \vec{B}$
$2 \sqrt{3} \times 10^{- 5} = I \times \sqrt{3} \times 10^{- 4} \times 10^{- 1} (M = I \times A)$
$I = 2 A$

AP EAMCET (20.04.2019) Shift-II

Moving Charges & Magnetism

153922 The resultant magnetic moment of three magnetic dipoles, each of the magnetic moment $M$ shown in the arrangement is

1

\sqrt{2} M

2

(\sqrt{2} + 1) M

3

(\sqrt{2} - 1) M

4

M

Explanation:

B Three magnetic moments of magnetic dipole, are shown in the figure;

$M_{1} = \sqrt{M_{A}^{2} + M_{B}^{2} + 2 M_{A} M_{B} \cos 90^{\circ}}$
$M_{1} = \sqrt{M^{2} + M^{2} + 0}$
$M_{1} = M \sqrt{2}$
The resultant of magnetic moment,
$M_{R} = M_{1} + M_{C}$
direction)
$(∵ M_{1} & M_{C} are in the same$
$M_{R} = M \sqrt{2} + M$
$M_{R} = (\sqrt{2} + 1) M$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153923 Magnetic field at the centre of a circular loop of area $A$ is $B$ . then the magnetic moment of the loop is ( $μ_{0}$ -permeability of the free space)

1

\frac{{BA}^{2}}{μ_{0} π}

2

\frac{BA \sqrt{A}}{μ_{0}}

3

\frac{BA \sqrt{A}}{μ_{0} π}

4

\frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}

Explanation:

D If $r$ is the radius of the circular loop then the area of the loop $(A) = π r^{2}$
$∴ r = \sqrt{\frac{A}{π}}$
The magnetic field at the centre of the loop,
$B = \frac{μ_{o}}{4 π} \times \frac{2 π I}{r} = \frac{μ_{o} I}{2 r}$
$I = \frac{2 Br}{μ_{o}}$
Magnetic moment,
$M = I \cdot A$
$M = \frac{2 Br}{μ_{o}} \cdot {(\frac{A}{π})}^{1 / 2}$
Putting the values of I and $r$ we get -
$M = \frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153920 Magnetic moment due to the motion of the electron in $n^{th}$ energy state of hydrogen atom is proportional to .............

1

n^{- 2}

2

n

3

n^{2}

4

n^{3}

Explanation:

B The magnetic moment due to the motion of electron in $n^{th}$ energy state of hydrogen atom is proportional to $n$ .
We know that,
$\frac{Magnetic moment}{Angular momentum} = \frac{e}{2 m}$
$\frac{M}{L} = \frac{e}{2 m}$
$M = \frac{L e}{2 m}$
Hence, $M \propto L$
Since, $L = \frac{nh}{2 π}$
So, $M \propto L \propto n$

AP EAMCET (21.04.2019) Shift-I

Moving Charges & Magnetism

153921 A coil in the shape of an equilateral triangle of side $2 cm$ is suspended from a vertex such that its hangs in a vertical plane between the poles of a permanent magnet producing a horizontal magnetic field of $100 \times 10^{- 3}$ T. The magnetic field is parallel to the plane of the coil. For the moment of couple acting on the coil to be $2 \sqrt{3} \times 10^{- 5} Nm$ , the current to be passed through the coil is

1

0.5 A

2

1 A

3

2 A

4

4 A

Explanation:

C Given :
$a = 2 \times 10^{- 2} m$
$A = \frac{\sqrt{3}}{4} \times a^{2}$
$A = \sqrt{3} \times 10^{- 4} m^{2}$
$∴ B = 100 \times 10^{- 3} T = 0.1 T$
$τ = 2 \sqrt{3} \times 10^{- 5} m$
We know that,
$\vec{τ} = \vec{M} \times \vec{B}$
$2 \sqrt{3} \times 10^{- 5} = I \times \sqrt{3} \times 10^{- 4} \times 10^{- 1} (M = I \times A)$
$I = 2 A$

AP EAMCET (20.04.2019) Shift-II

Moving Charges & Magnetism

153922 The resultant magnetic moment of three magnetic dipoles, each of the magnetic moment $M$ shown in the arrangement is

1

\sqrt{2} M

2

(\sqrt{2} + 1) M

3

(\sqrt{2} - 1) M

4

M

Explanation:

B Three magnetic moments of magnetic dipole, are shown in the figure;

$M_{1} = \sqrt{M_{A}^{2} + M_{B}^{2} + 2 M_{A} M_{B} \cos 90^{\circ}}$
$M_{1} = \sqrt{M^{2} + M^{2} + 0}$
$M_{1} = M \sqrt{2}$
The resultant of magnetic moment,
$M_{R} = M_{1} + M_{C}$
direction)
$(∵ M_{1} & M_{C} are in the same$
$M_{R} = M \sqrt{2} + M$
$M_{R} = (\sqrt{2} + 1) M$

AP EAMCET (23.04.2019) Shift-I

Moving Charges & Magnetism

153923 Magnetic field at the centre of a circular loop of area $A$ is $B$ . then the magnetic moment of the loop is ( $μ_{0}$ -permeability of the free space)

1

\frac{{BA}^{2}}{μ_{0} π}

2

\frac{BA \sqrt{A}}{μ_{0}}

3

\frac{BA \sqrt{A}}{μ_{0} π}

4

\frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}

Explanation:

D If $r$ is the radius of the circular loop then the area of the loop $(A) = π r^{2}$
$∴ r = \sqrt{\frac{A}{π}}$
The magnetic field at the centre of the loop,
$B = \frac{μ_{o}}{4 π} \times \frac{2 π I}{r} = \frac{μ_{o} I}{2 r}$
$I = \frac{2 Br}{μ_{o}}$
Magnetic moment,
$M = I \cdot A$
$M = \frac{2 Br}{μ_{o}} \cdot {(\frac{A}{π})}^{1 / 2}$
Putting the values of I and $r$ we get -
$M = \frac{2 BA \sqrt{A}}{μ_{0} \sqrt{π}}$

AP EAMCET (23.04.2019) Shift-I

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