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PHXII05:MAGNETISM and MATTER

360605 A magnet of magnetic moment $M$ is in the form of a quadrant of a circle. If it is straightened, its new magnetic moment will be

1

\frac{M π}{\sqrt{2}}

2

\frac{M}{\sqrt{2}}

3

\frac{\sqrt{2} M}{π}

4

\frac{M π}{2 \sqrt{2}}

Explanation:

Let $m$ be the pole strength of the magnet. Let $M$ and $M^{'}$ be the magnetic moments when the magnet is in arc shape and straight shape respectively. Let $θ$ be the angle subtents by the arc at the centre.
supporting img
$M = m 2 R \sin (\frac{θ}{2}) (1)$
$2 R \sin (\frac{θ}{2})$ is the effective length of the magnet.
The arc length of the magnet is $R θ$ . When the magnet is made straight then the magnetic moment is
$M^{'} = m l = m R θ (2)$
From eqns $1 & 2$
$\frac{M}{M^{'}} = \frac{2 \sin (\frac{θ}{2})}{θ}$
Given that $θ = \frac{π}{2}$
$\frac{M}{M^{'}} = \frac{2 R \sin (\frac{π}{4})}{\frac{π}{2}} \Rightarrow M^{'} = \frac{π M}{2 \sqrt{2}}$

PHXII05:MAGNETISM and MATTER

360606 If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be

1

\frac{2 m}{e J}

2

\frac{e J}{2 m}

3

\frac{e J}{m}

4

\frac{e J}{4 m}

Explanation:

As we know for circulating electron magnetic moment
$M = i π r^{2} = \frac{e ω π r^{2}}{2 π} = \frac{1}{2} e v r (1)$
and angular momentum $J = m v r (2)$
From equation (1) and (2), $M = \frac{e J}{2 m}$

PHXII05:MAGNETISM and MATTER

360607 A magnet of pole strength $m$ and length $l$ is broken into two pieces. The pole strength of each piece is

1

m

2

m / 2

3

2 m

4

m / 4

PHXII05:MAGNETISM and MATTER

360608 Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is
supporting img

1

2 M

2 Zero

3

\frac{3 M}{2}

4

M \sqrt{3}

Explanation:

The resultant magnetic moment can be calculated as follows.
supporting img
$\begin{aligned} \overset{―}{M_{1}} + \overset{―}{M_{3}} = \overset{―}{M_{2}} \\ \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} = (\overset{―}{M_{2}}) + \overset{―}{M_{2}} = 2 \overset{―}{M_{2}} \\ | \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} | = | 2 \overset{―}{M_{2}} | = 2 M \end{aligned}$

PHXII05:MAGNETISM and MATTER

360605 A magnet of magnetic moment $M$ is in the form of a quadrant of a circle. If it is straightened, its new magnetic moment will be

1

\frac{M π}{\sqrt{2}}

2

\frac{M}{\sqrt{2}}

3

\frac{\sqrt{2} M}{π}

4

\frac{M π}{2 \sqrt{2}}

Explanation:

Let $m$ be the pole strength of the magnet. Let $M$ and $M^{'}$ be the magnetic moments when the magnet is in arc shape and straight shape respectively. Let $θ$ be the angle subtents by the arc at the centre.
supporting img
$M = m 2 R \sin (\frac{θ}{2}) (1)$
$2 R \sin (\frac{θ}{2})$ is the effective length of the magnet.
The arc length of the magnet is $R θ$ . When the magnet is made straight then the magnetic moment is
$M^{'} = m l = m R θ (2)$
From eqns $1 & 2$
$\frac{M}{M^{'}} = \frac{2 \sin (\frac{θ}{2})}{θ}$
Given that $θ = \frac{π}{2}$
$\frac{M}{M^{'}} = \frac{2 R \sin (\frac{π}{4})}{\frac{π}{2}} \Rightarrow M^{'} = \frac{π M}{2 \sqrt{2}}$

PHXII05:MAGNETISM and MATTER

360606 If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be

1

\frac{2 m}{e J}

2

\frac{e J}{2 m}

3

\frac{e J}{m}

4

\frac{e J}{4 m}

Explanation:

As we know for circulating electron magnetic moment
$M = i π r^{2} = \frac{e ω π r^{2}}{2 π} = \frac{1}{2} e v r (1)$
and angular momentum $J = m v r (2)$
From equation (1) and (2), $M = \frac{e J}{2 m}$

PHXII05:MAGNETISM and MATTER

360607 A magnet of pole strength $m$ and length $l$ is broken into two pieces. The pole strength of each piece is

1

m

2

m / 2

3

2 m

4

m / 4

PHXII05:MAGNETISM and MATTER

360608 Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is
supporting img

1

2 M

2 Zero

3

\frac{3 M}{2}

4

M \sqrt{3}

Explanation:

The resultant magnetic moment can be calculated as follows.
supporting img
$\begin{aligned} \overset{―}{M_{1}} + \overset{―}{M_{3}} = \overset{―}{M_{2}} \\ \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} = (\overset{―}{M_{2}}) + \overset{―}{M_{2}} = 2 \overset{―}{M_{2}} \\ | \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} | = | 2 \overset{―}{M_{2}} | = 2 M \end{aligned}$

PHXII05:MAGNETISM and MATTER

360605 A magnet of magnetic moment $M$ is in the form of a quadrant of a circle. If it is straightened, its new magnetic moment will be

1

\frac{M π}{\sqrt{2}}

2

\frac{M}{\sqrt{2}}

3

\frac{\sqrt{2} M}{π}

4

\frac{M π}{2 \sqrt{2}}

Explanation:

Let $m$ be the pole strength of the magnet. Let $M$ and $M^{'}$ be the magnetic moments when the magnet is in arc shape and straight shape respectively. Let $θ$ be the angle subtents by the arc at the centre.
supporting img
$M = m 2 R \sin (\frac{θ}{2}) (1)$
$2 R \sin (\frac{θ}{2})$ is the effective length of the magnet.
The arc length of the magnet is $R θ$ . When the magnet is made straight then the magnetic moment is
$M^{'} = m l = m R θ (2)$
From eqns $1 & 2$
$\frac{M}{M^{'}} = \frac{2 \sin (\frac{θ}{2})}{θ}$
Given that $θ = \frac{π}{2}$
$\frac{M}{M^{'}} = \frac{2 R \sin (\frac{π}{4})}{\frac{π}{2}} \Rightarrow M^{'} = \frac{π M}{2 \sqrt{2}}$

PHXII05:MAGNETISM and MATTER

360606 If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be

1

\frac{2 m}{e J}

2

\frac{e J}{2 m}

3

\frac{e J}{m}

4

\frac{e J}{4 m}

Explanation:

As we know for circulating electron magnetic moment
$M = i π r^{2} = \frac{e ω π r^{2}}{2 π} = \frac{1}{2} e v r (1)$
and angular momentum $J = m v r (2)$
From equation (1) and (2), $M = \frac{e J}{2 m}$

PHXII05:MAGNETISM and MATTER

360607 A magnet of pole strength $m$ and length $l$ is broken into two pieces. The pole strength of each piece is

1

m

2

m / 2

3

2 m

4

m / 4

PHXII05:MAGNETISM and MATTER

360608 Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is
supporting img

1

2 M

2 Zero

3

\frac{3 M}{2}

4

M \sqrt{3}

Explanation:

The resultant magnetic moment can be calculated as follows.
supporting img
$\begin{aligned} \overset{―}{M_{1}} + \overset{―}{M_{3}} = \overset{―}{M_{2}} \\ \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} = (\overset{―}{M_{2}}) + \overset{―}{M_{2}} = 2 \overset{―}{M_{2}} \\ | \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} | = | 2 \overset{―}{M_{2}} | = 2 M \end{aligned}$

PHXII05:MAGNETISM and MATTER

360605 A magnet of magnetic moment $M$ is in the form of a quadrant of a circle. If it is straightened, its new magnetic moment will be

1

\frac{M π}{\sqrt{2}}

2

\frac{M}{\sqrt{2}}

3

\frac{\sqrt{2} M}{π}

4

\frac{M π}{2 \sqrt{2}}

Explanation:

Let $m$ be the pole strength of the magnet. Let $M$ and $M^{'}$ be the magnetic moments when the magnet is in arc shape and straight shape respectively. Let $θ$ be the angle subtents by the arc at the centre.
supporting img
$M = m 2 R \sin (\frac{θ}{2}) (1)$
$2 R \sin (\frac{θ}{2})$ is the effective length of the magnet.
The arc length of the magnet is $R θ$ . When the magnet is made straight then the magnetic moment is
$M^{'} = m l = m R θ (2)$
From eqns $1 & 2$
$\frac{M}{M^{'}} = \frac{2 \sin (\frac{θ}{2})}{θ}$
Given that $θ = \frac{π}{2}$
$\frac{M}{M^{'}} = \frac{2 R \sin (\frac{π}{4})}{\frac{π}{2}} \Rightarrow M^{'} = \frac{π M}{2 \sqrt{2}}$

PHXII05:MAGNETISM and MATTER

360606 If the angular momentum of an electron is $\vec{J}$ then the magnitude of the magnetic moment will be

1

\frac{2 m}{e J}

2

\frac{e J}{2 m}

3

\frac{e J}{m}

4

\frac{e J}{4 m}

Explanation:

As we know for circulating electron magnetic moment
$M = i π r^{2} = \frac{e ω π r^{2}}{2 π} = \frac{1}{2} e v r (1)$
and angular momentum $J = m v r (2)$
From equation (1) and (2), $M = \frac{e J}{2 m}$

PHXII05:MAGNETISM and MATTER

360607 A magnet of pole strength $m$ and length $l$ is broken into two pieces. The pole strength of each piece is

1

m

2

m / 2

3

2 m

4

m / 4

PHXII05:MAGNETISM and MATTER

360608 Three identical bar magnets each of magnetic moment $M$ are placed in the form of an equilateral triangle as shown. The net magnetic moment of the system is
supporting img

1

2 M

2 Zero

3

\frac{3 M}{2}

4

M \sqrt{3}

Explanation:

The resultant magnetic moment can be calculated as follows.
supporting img
$\begin{aligned} \overset{―}{M_{1}} + \overset{―}{M_{3}} = \overset{―}{M_{2}} \\ \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} = (\overset{―}{M_{2}}) + \overset{―}{M_{2}} = 2 \overset{―}{M_{2}} \\ | \overset{―}{M_{1}} + \overset{―}{M_{2}} + \overset{―}{M_{3}} | = | 2 \overset{―}{M_{2}} | = 2 M \end{aligned}$