153956
The magnetised wire moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then, its magnetic moment is
1 $\frac{2 \mathrm{M}}{\pi}$
2 $2 \mathrm{M}$
3 $\frac{M}{\pi}$
4 zero
Explanation:
A Assume length of wire \(=l\) Given, $\text { Magnetic moment }=\mathrm{M}$ Length of wire $=l$ After bent into semicircle Circumference of semicircle $=\pi \mathrm{r}=l$ $\mathrm{r}=\frac{l}{\pi}$ Magnetic moment $\mathrm{M}=\mathrm{m} l \Rightarrow \frac{\mathrm{M}}{l}=\mathrm{m}$ Length between two poles after bent into semicircle $=$ $2 \mathrm{r}$ $\mathrm{M}^{\prime}=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{\mathrm{M}}{l} \times 2 \mathrm{r}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{l} \quad(\because l=\pi \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{\pi \mathrm{r}}=\frac{2 \mathrm{M}}{\pi}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{M}}{\pi}$
AP EAMCET-04.07.2022
Moving Charges & Magnetism
153957
A bar magnet of moment of inertia $49 \times 10^{-2}$ $\mathrm{kg}-\mathrm{m}^{2}$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
1 $350 \mathrm{~A}-\mathrm{m}^{2}$
2 $490 \mathrm{~A}-\mathrm{m}^{2}$
3 $3300 \mathrm{~A}-\mathrm{m}^{2}$
4 $5000 \mathrm{~A}-\mathrm{m}^{2}$
Explanation:
D Given, Moment of inertia $\mathrm{M}=49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}$ Magnetic field $=0.5 \times 10^{-4} \mathrm{~T}$ Time period $=8.8 \mathrm{sec}$ $\because$ Time period of vibrational magnetometer is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}}$ $8.8=2 \pi \sqrt{\frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}}$ Squaring both side $(8.8)^{2}=(2 \pi)^{2} \frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}$ $\mathrm{M}=\frac{(2 \pi)^{2} \times 49 \times 10^{-2}}{(8.8)^{2} \times 0.5 \times 10^{-4}}=5000 \mathrm{~A}-\mathrm{m}^{2}$
EAMCET-2007
Moving Charges & Magnetism
153961
A bar magnet of magnetic moment $20 \mathrm{~J} / \mathrm{T}$ lies aligned with the direction of a uniform magnetic field of 0.25T. The amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction is
1 $0.10 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $0.3 \mathrm{~J}$
4 $5.0 \mathrm{~J}$
Explanation:
D Given, Magnetic moment $(\mathrm{M})=20 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.25 \mathrm{~T}$ The amount of work, $\mathrm{W}=\mathrm{M} \times \mathrm{B}$ $\mathrm{W}=20 \times 0.25$ $\mathrm{~W}=5 \mathrm{~J}$
TS EAMCET(Medical)-2015
Moving Charges & Magnetism
153963
The magnetic field at the centre of as circular loop of area $A$ is $B$. The magnetic moment of the loop will be
B The magnetic field due to a current carrying circular loop at the centre. $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment of the loop $\mathrm{M}=\mathrm{IA}$ From equation (i) we get- $\mathrm{M}=\frac{2 \mathrm{BR}}{\mu_{0}} \cdot \mathrm{A}$ We know that, area of circular loop $\mathrm{A}=\pi \mathrm{R}^{2}$ $\mathrm{R}=\sqrt{\frac{\mathrm{A}}{\pi}}$ Hence, $\mathrm{M}=\frac{2 \mathrm{BA} \sqrt{\mathrm{A}}}{\mu_{0} \sqrt{\pi}}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_{0} \pi^{1 / 2}}$
Assam CEE-2016
Moving Charges & Magnetism
153964
A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v_{1}{ }^{\prime}$ where $v\lt\lt$ speed of light. The ratio of magnetic moment of the particle to its angular momentum is
1 $\frac{\mathrm{q}}{2 \mathrm{M}}$
2 $\frac{M}{2 q}$
3 $\frac{\mathrm{q}}{\mathrm{M}}$
4 $\frac{\mathrm{q}}{4 \mathrm{M}}$
Explanation:
A The angular momentum $\mathrm{L}$ of the particle is $\mathrm{L}=\mathrm{Mr}^{2} \omega$ Frequency $(f)=\frac{\omega}{2 \pi}$ $\mathrm{i}=\mathrm{q} \times \mathrm{f}=\frac{\omega \mathrm{q}}{2 \pi}$ $\mathrm{m}=\mathrm{iA}=\frac{\omega \mathrm{q}}{2 \pi} \times \pi \mathrm{r}^{2}$ So, Angular momentum, $\frac{\mathrm{m}}{\mathrm{L}}=\frac{\omega \mathrm{qr}^{2}}{2 \mathrm{Mr}^{2} \omega}=\frac{\mathrm{q}}{2 \mathrm{M}}$
153956
The magnetised wire moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then, its magnetic moment is
1 $\frac{2 \mathrm{M}}{\pi}$
2 $2 \mathrm{M}$
3 $\frac{M}{\pi}$
4 zero
Explanation:
A Assume length of wire \(=l\) Given, $\text { Magnetic moment }=\mathrm{M}$ Length of wire $=l$ After bent into semicircle Circumference of semicircle $=\pi \mathrm{r}=l$ $\mathrm{r}=\frac{l}{\pi}$ Magnetic moment $\mathrm{M}=\mathrm{m} l \Rightarrow \frac{\mathrm{M}}{l}=\mathrm{m}$ Length between two poles after bent into semicircle $=$ $2 \mathrm{r}$ $\mathrm{M}^{\prime}=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{\mathrm{M}}{l} \times 2 \mathrm{r}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{l} \quad(\because l=\pi \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{\pi \mathrm{r}}=\frac{2 \mathrm{M}}{\pi}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{M}}{\pi}$
AP EAMCET-04.07.2022
Moving Charges & Magnetism
153957
A bar magnet of moment of inertia $49 \times 10^{-2}$ $\mathrm{kg}-\mathrm{m}^{2}$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
1 $350 \mathrm{~A}-\mathrm{m}^{2}$
2 $490 \mathrm{~A}-\mathrm{m}^{2}$
3 $3300 \mathrm{~A}-\mathrm{m}^{2}$
4 $5000 \mathrm{~A}-\mathrm{m}^{2}$
Explanation:
D Given, Moment of inertia $\mathrm{M}=49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}$ Magnetic field $=0.5 \times 10^{-4} \mathrm{~T}$ Time period $=8.8 \mathrm{sec}$ $\because$ Time period of vibrational magnetometer is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}}$ $8.8=2 \pi \sqrt{\frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}}$ Squaring both side $(8.8)^{2}=(2 \pi)^{2} \frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}$ $\mathrm{M}=\frac{(2 \pi)^{2} \times 49 \times 10^{-2}}{(8.8)^{2} \times 0.5 \times 10^{-4}}=5000 \mathrm{~A}-\mathrm{m}^{2}$
EAMCET-2007
Moving Charges & Magnetism
153961
A bar magnet of magnetic moment $20 \mathrm{~J} / \mathrm{T}$ lies aligned with the direction of a uniform magnetic field of 0.25T. The amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction is
1 $0.10 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $0.3 \mathrm{~J}$
4 $5.0 \mathrm{~J}$
Explanation:
D Given, Magnetic moment $(\mathrm{M})=20 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.25 \mathrm{~T}$ The amount of work, $\mathrm{W}=\mathrm{M} \times \mathrm{B}$ $\mathrm{W}=20 \times 0.25$ $\mathrm{~W}=5 \mathrm{~J}$
TS EAMCET(Medical)-2015
Moving Charges & Magnetism
153963
The magnetic field at the centre of as circular loop of area $A$ is $B$. The magnetic moment of the loop will be
B The magnetic field due to a current carrying circular loop at the centre. $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment of the loop $\mathrm{M}=\mathrm{IA}$ From equation (i) we get- $\mathrm{M}=\frac{2 \mathrm{BR}}{\mu_{0}} \cdot \mathrm{A}$ We know that, area of circular loop $\mathrm{A}=\pi \mathrm{R}^{2}$ $\mathrm{R}=\sqrt{\frac{\mathrm{A}}{\pi}}$ Hence, $\mathrm{M}=\frac{2 \mathrm{BA} \sqrt{\mathrm{A}}}{\mu_{0} \sqrt{\pi}}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_{0} \pi^{1 / 2}}$
Assam CEE-2016
Moving Charges & Magnetism
153964
A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v_{1}{ }^{\prime}$ where $v\lt\lt$ speed of light. The ratio of magnetic moment of the particle to its angular momentum is
1 $\frac{\mathrm{q}}{2 \mathrm{M}}$
2 $\frac{M}{2 q}$
3 $\frac{\mathrm{q}}{\mathrm{M}}$
4 $\frac{\mathrm{q}}{4 \mathrm{M}}$
Explanation:
A The angular momentum $\mathrm{L}$ of the particle is $\mathrm{L}=\mathrm{Mr}^{2} \omega$ Frequency $(f)=\frac{\omega}{2 \pi}$ $\mathrm{i}=\mathrm{q} \times \mathrm{f}=\frac{\omega \mathrm{q}}{2 \pi}$ $\mathrm{m}=\mathrm{iA}=\frac{\omega \mathrm{q}}{2 \pi} \times \pi \mathrm{r}^{2}$ So, Angular momentum, $\frac{\mathrm{m}}{\mathrm{L}}=\frac{\omega \mathrm{qr}^{2}}{2 \mathrm{Mr}^{2} \omega}=\frac{\mathrm{q}}{2 \mathrm{M}}$
153956
The magnetised wire moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then, its magnetic moment is
1 $\frac{2 \mathrm{M}}{\pi}$
2 $2 \mathrm{M}$
3 $\frac{M}{\pi}$
4 zero
Explanation:
A Assume length of wire \(=l\) Given, $\text { Magnetic moment }=\mathrm{M}$ Length of wire $=l$ After bent into semicircle Circumference of semicircle $=\pi \mathrm{r}=l$ $\mathrm{r}=\frac{l}{\pi}$ Magnetic moment $\mathrm{M}=\mathrm{m} l \Rightarrow \frac{\mathrm{M}}{l}=\mathrm{m}$ Length between two poles after bent into semicircle $=$ $2 \mathrm{r}$ $\mathrm{M}^{\prime}=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{\mathrm{M}}{l} \times 2 \mathrm{r}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{l} \quad(\because l=\pi \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{\pi \mathrm{r}}=\frac{2 \mathrm{M}}{\pi}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{M}}{\pi}$
AP EAMCET-04.07.2022
Moving Charges & Magnetism
153957
A bar magnet of moment of inertia $49 \times 10^{-2}$ $\mathrm{kg}-\mathrm{m}^{2}$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
1 $350 \mathrm{~A}-\mathrm{m}^{2}$
2 $490 \mathrm{~A}-\mathrm{m}^{2}$
3 $3300 \mathrm{~A}-\mathrm{m}^{2}$
4 $5000 \mathrm{~A}-\mathrm{m}^{2}$
Explanation:
D Given, Moment of inertia $\mathrm{M}=49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}$ Magnetic field $=0.5 \times 10^{-4} \mathrm{~T}$ Time period $=8.8 \mathrm{sec}$ $\because$ Time period of vibrational magnetometer is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}}$ $8.8=2 \pi \sqrt{\frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}}$ Squaring both side $(8.8)^{2}=(2 \pi)^{2} \frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}$ $\mathrm{M}=\frac{(2 \pi)^{2} \times 49 \times 10^{-2}}{(8.8)^{2} \times 0.5 \times 10^{-4}}=5000 \mathrm{~A}-\mathrm{m}^{2}$
EAMCET-2007
Moving Charges & Magnetism
153961
A bar magnet of magnetic moment $20 \mathrm{~J} / \mathrm{T}$ lies aligned with the direction of a uniform magnetic field of 0.25T. The amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction is
1 $0.10 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $0.3 \mathrm{~J}$
4 $5.0 \mathrm{~J}$
Explanation:
D Given, Magnetic moment $(\mathrm{M})=20 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.25 \mathrm{~T}$ The amount of work, $\mathrm{W}=\mathrm{M} \times \mathrm{B}$ $\mathrm{W}=20 \times 0.25$ $\mathrm{~W}=5 \mathrm{~J}$
TS EAMCET(Medical)-2015
Moving Charges & Magnetism
153963
The magnetic field at the centre of as circular loop of area $A$ is $B$. The magnetic moment of the loop will be
B The magnetic field due to a current carrying circular loop at the centre. $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment of the loop $\mathrm{M}=\mathrm{IA}$ From equation (i) we get- $\mathrm{M}=\frac{2 \mathrm{BR}}{\mu_{0}} \cdot \mathrm{A}$ We know that, area of circular loop $\mathrm{A}=\pi \mathrm{R}^{2}$ $\mathrm{R}=\sqrt{\frac{\mathrm{A}}{\pi}}$ Hence, $\mathrm{M}=\frac{2 \mathrm{BA} \sqrt{\mathrm{A}}}{\mu_{0} \sqrt{\pi}}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_{0} \pi^{1 / 2}}$
Assam CEE-2016
Moving Charges & Magnetism
153964
A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v_{1}{ }^{\prime}$ where $v\lt\lt$ speed of light. The ratio of magnetic moment of the particle to its angular momentum is
1 $\frac{\mathrm{q}}{2 \mathrm{M}}$
2 $\frac{M}{2 q}$
3 $\frac{\mathrm{q}}{\mathrm{M}}$
4 $\frac{\mathrm{q}}{4 \mathrm{M}}$
Explanation:
A The angular momentum $\mathrm{L}$ of the particle is $\mathrm{L}=\mathrm{Mr}^{2} \omega$ Frequency $(f)=\frac{\omega}{2 \pi}$ $\mathrm{i}=\mathrm{q} \times \mathrm{f}=\frac{\omega \mathrm{q}}{2 \pi}$ $\mathrm{m}=\mathrm{iA}=\frac{\omega \mathrm{q}}{2 \pi} \times \pi \mathrm{r}^{2}$ So, Angular momentum, $\frac{\mathrm{m}}{\mathrm{L}}=\frac{\omega \mathrm{qr}^{2}}{2 \mathrm{Mr}^{2} \omega}=\frac{\mathrm{q}}{2 \mathrm{M}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Moving Charges & Magnetism
153956
The magnetised wire moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then, its magnetic moment is
1 $\frac{2 \mathrm{M}}{\pi}$
2 $2 \mathrm{M}$
3 $\frac{M}{\pi}$
4 zero
Explanation:
A Assume length of wire \(=l\) Given, $\text { Magnetic moment }=\mathrm{M}$ Length of wire $=l$ After bent into semicircle Circumference of semicircle $=\pi \mathrm{r}=l$ $\mathrm{r}=\frac{l}{\pi}$ Magnetic moment $\mathrm{M}=\mathrm{m} l \Rightarrow \frac{\mathrm{M}}{l}=\mathrm{m}$ Length between two poles after bent into semicircle $=$ $2 \mathrm{r}$ $\mathrm{M}^{\prime}=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{\mathrm{M}}{l} \times 2 \mathrm{r}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{l} \quad(\because l=\pi \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{\pi \mathrm{r}}=\frac{2 \mathrm{M}}{\pi}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{M}}{\pi}$
AP EAMCET-04.07.2022
Moving Charges & Magnetism
153957
A bar magnet of moment of inertia $49 \times 10^{-2}$ $\mathrm{kg}-\mathrm{m}^{2}$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
1 $350 \mathrm{~A}-\mathrm{m}^{2}$
2 $490 \mathrm{~A}-\mathrm{m}^{2}$
3 $3300 \mathrm{~A}-\mathrm{m}^{2}$
4 $5000 \mathrm{~A}-\mathrm{m}^{2}$
Explanation:
D Given, Moment of inertia $\mathrm{M}=49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}$ Magnetic field $=0.5 \times 10^{-4} \mathrm{~T}$ Time period $=8.8 \mathrm{sec}$ $\because$ Time period of vibrational magnetometer is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}}$ $8.8=2 \pi \sqrt{\frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}}$ Squaring both side $(8.8)^{2}=(2 \pi)^{2} \frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}$ $\mathrm{M}=\frac{(2 \pi)^{2} \times 49 \times 10^{-2}}{(8.8)^{2} \times 0.5 \times 10^{-4}}=5000 \mathrm{~A}-\mathrm{m}^{2}$
EAMCET-2007
Moving Charges & Magnetism
153961
A bar magnet of magnetic moment $20 \mathrm{~J} / \mathrm{T}$ lies aligned with the direction of a uniform magnetic field of 0.25T. The amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction is
1 $0.10 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $0.3 \mathrm{~J}$
4 $5.0 \mathrm{~J}$
Explanation:
D Given, Magnetic moment $(\mathrm{M})=20 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.25 \mathrm{~T}$ The amount of work, $\mathrm{W}=\mathrm{M} \times \mathrm{B}$ $\mathrm{W}=20 \times 0.25$ $\mathrm{~W}=5 \mathrm{~J}$
TS EAMCET(Medical)-2015
Moving Charges & Magnetism
153963
The magnetic field at the centre of as circular loop of area $A$ is $B$. The magnetic moment of the loop will be
B The magnetic field due to a current carrying circular loop at the centre. $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment of the loop $\mathrm{M}=\mathrm{IA}$ From equation (i) we get- $\mathrm{M}=\frac{2 \mathrm{BR}}{\mu_{0}} \cdot \mathrm{A}$ We know that, area of circular loop $\mathrm{A}=\pi \mathrm{R}^{2}$ $\mathrm{R}=\sqrt{\frac{\mathrm{A}}{\pi}}$ Hence, $\mathrm{M}=\frac{2 \mathrm{BA} \sqrt{\mathrm{A}}}{\mu_{0} \sqrt{\pi}}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_{0} \pi^{1 / 2}}$
Assam CEE-2016
Moving Charges & Magnetism
153964
A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v_{1}{ }^{\prime}$ where $v\lt\lt$ speed of light. The ratio of magnetic moment of the particle to its angular momentum is
1 $\frac{\mathrm{q}}{2 \mathrm{M}}$
2 $\frac{M}{2 q}$
3 $\frac{\mathrm{q}}{\mathrm{M}}$
4 $\frac{\mathrm{q}}{4 \mathrm{M}}$
Explanation:
A The angular momentum $\mathrm{L}$ of the particle is $\mathrm{L}=\mathrm{Mr}^{2} \omega$ Frequency $(f)=\frac{\omega}{2 \pi}$ $\mathrm{i}=\mathrm{q} \times \mathrm{f}=\frac{\omega \mathrm{q}}{2 \pi}$ $\mathrm{m}=\mathrm{iA}=\frac{\omega \mathrm{q}}{2 \pi} \times \pi \mathrm{r}^{2}$ So, Angular momentum, $\frac{\mathrm{m}}{\mathrm{L}}=\frac{\omega \mathrm{qr}^{2}}{2 \mathrm{Mr}^{2} \omega}=\frac{\mathrm{q}}{2 \mathrm{M}}$
153956
The magnetised wire moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then, its magnetic moment is
1 $\frac{2 \mathrm{M}}{\pi}$
2 $2 \mathrm{M}$
3 $\frac{M}{\pi}$
4 zero
Explanation:
A Assume length of wire \(=l\) Given, $\text { Magnetic moment }=\mathrm{M}$ Length of wire $=l$ After bent into semicircle Circumference of semicircle $=\pi \mathrm{r}=l$ $\mathrm{r}=\frac{l}{\pi}$ Magnetic moment $\mathrm{M}=\mathrm{m} l \Rightarrow \frac{\mathrm{M}}{l}=\mathrm{m}$ Length between two poles after bent into semicircle $=$ $2 \mathrm{r}$ $\mathrm{M}^{\prime}=\mathrm{m}(2 \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{\mathrm{M}}{l} \times 2 \mathrm{r}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{l} \quad(\because l=\pi \mathrm{r})$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{Mr}}{\pi \mathrm{r}}=\frac{2 \mathrm{M}}{\pi}$ $\mathrm{M}^{\prime} =\frac{2 \mathrm{M}}{\pi}$
AP EAMCET-04.07.2022
Moving Charges & Magnetism
153957
A bar magnet of moment of inertia $49 \times 10^{-2}$ $\mathrm{kg}-\mathrm{m}^{2}$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is 8.8 s. The magnetic moment of the bar magnet is
1 $350 \mathrm{~A}-\mathrm{m}^{2}$
2 $490 \mathrm{~A}-\mathrm{m}^{2}$
3 $3300 \mathrm{~A}-\mathrm{m}^{2}$
4 $5000 \mathrm{~A}-\mathrm{m}^{2}$
Explanation:
D Given, Moment of inertia $\mathrm{M}=49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}$ Magnetic field $=0.5 \times 10^{-4} \mathrm{~T}$ Time period $=8.8 \mathrm{sec}$ $\because$ Time period of vibrational magnetometer is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}}$ $8.8=2 \pi \sqrt{\frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}}$ Squaring both side $(8.8)^{2}=(2 \pi)^{2} \frac{49 \times 10^{-2}}{\mathrm{M} \times 0.5 \times 10^{-4}}$ $\mathrm{M}=\frac{(2 \pi)^{2} \times 49 \times 10^{-2}}{(8.8)^{2} \times 0.5 \times 10^{-4}}=5000 \mathrm{~A}-\mathrm{m}^{2}$
EAMCET-2007
Moving Charges & Magnetism
153961
A bar magnet of magnetic moment $20 \mathrm{~J} / \mathrm{T}$ lies aligned with the direction of a uniform magnetic field of 0.25T. The amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction is
1 $0.10 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $0.3 \mathrm{~J}$
4 $5.0 \mathrm{~J}$
Explanation:
D Given, Magnetic moment $(\mathrm{M})=20 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.25 \mathrm{~T}$ The amount of work, $\mathrm{W}=\mathrm{M} \times \mathrm{B}$ $\mathrm{W}=20 \times 0.25$ $\mathrm{~W}=5 \mathrm{~J}$
TS EAMCET(Medical)-2015
Moving Charges & Magnetism
153963
The magnetic field at the centre of as circular loop of area $A$ is $B$. The magnetic moment of the loop will be
B The magnetic field due to a current carrying circular loop at the centre. $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{R}}$ Magnetic moment of the loop $\mathrm{M}=\mathrm{IA}$ From equation (i) we get- $\mathrm{M}=\frac{2 \mathrm{BR}}{\mu_{0}} \cdot \mathrm{A}$ We know that, area of circular loop $\mathrm{A}=\pi \mathrm{R}^{2}$ $\mathrm{R}=\sqrt{\frac{\mathrm{A}}{\pi}}$ Hence, $\mathrm{M}=\frac{2 \mathrm{BA} \sqrt{\mathrm{A}}}{\mu_{0} \sqrt{\pi}}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_{0} \pi^{1 / 2}}$
Assam CEE-2016
Moving Charges & Magnetism
153964
A particle of mass $M$ and charge $q$ is moving in a circle of radius $R$ with speed $v_{1}{ }^{\prime}$ where $v\lt\lt$ speed of light. The ratio of magnetic moment of the particle to its angular momentum is
1 $\frac{\mathrm{q}}{2 \mathrm{M}}$
2 $\frac{M}{2 q}$
3 $\frac{\mathrm{q}}{\mathrm{M}}$
4 $\frac{\mathrm{q}}{4 \mathrm{M}}$
Explanation:
A The angular momentum $\mathrm{L}$ of the particle is $\mathrm{L}=\mathrm{Mr}^{2} \omega$ Frequency $(f)=\frac{\omega}{2 \pi}$ $\mathrm{i}=\mathrm{q} \times \mathrm{f}=\frac{\omega \mathrm{q}}{2 \pi}$ $\mathrm{m}=\mathrm{iA}=\frac{\omega \mathrm{q}}{2 \pi} \times \pi \mathrm{r}^{2}$ So, Angular momentum, $\frac{\mathrm{m}}{\mathrm{L}}=\frac{\omega \mathrm{qr}^{2}}{2 \mathrm{Mr}^{2} \omega}=\frac{\mathrm{q}}{2 \mathrm{M}}$
