D Given, Radius of circular loop $(\mathrm{r})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Area of loop $(\mathrm{A})=\pi \mathrm{r}^{2}=\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}^{2}$ Current in the loop $(\mathrm{I})=0.1 \mathrm{~A}$ We know that, magnetic moment $(\mathrm{M})=\mathrm{I}$.A $\mathrm{M}=0.1 \times \pi\left(5 \times 10^{-2}\right)^{2}$ $\mathrm{M}=7.85 \times 10^{-4} \mathrm{~A}-\mathrm{m}^{2}$
CG PET- 2005
Moving Charges & Magnetism
153897
A magnet of magnetic moment $6 \mathrm{JT}^{-1}$ is aligned in the direction of magnetic field of $0.3 \mathrm{~T}$. The net work done to bring the magnet normal to the magnetic field is
1 $2 \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $1.8 \mathrm{~J}$
4 $2.4 \mathrm{~J}$
Explanation:
C Work required to turn the dipole is given by $\mathrm{W}=\mathrm{MB}\left(\cos \theta_{\mathrm{i}}-\cos \theta_{\mathrm{f}}\right)$ Here, $\theta_{i}=0^{\circ} \quad$ and $\theta_{\mathrm{f}}=90^{\circ}$ $\mathrm{W}=6 \times 0.3\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=1.8(1-0)$ $\mathrm{W}=1.8 \mathrm{~J}$
AP EAMCET-07.09.2021
Moving Charges & Magnetism
153903
Magnetic potential at any point on equatorial line of the magnetic dipole is
D Magnetic potential, $\mathrm{V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos \theta}{r^{2}}$ $\mathrm{~V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos 90^{\circ}}{\mathrm{r}^{2}}$ $\mathrm{~V}=0$
AP EAMCET (Medical)-07.10.2020
Moving Charges & Magnetism
153905
The magnetic moment of current (I) carrying circular coil of radius ( $r$ ) and number of turns (n) varies as:
1 $1 / \mathrm{r}^{2}$
2 $1 / \mathrm{r}$
3 $r$
4 $\mathrm{r}^{2}$
Explanation:
D We know that, the magnetic movement of a current carrying circular coil is given as- $\mathrm{M}=\mathrm{nIA}=\mathrm{nI} \pi \mathrm{r}^{2}$ Hence, magnetic moment of circular coil carrying current (I) of radius $r$ and number turns $\mathrm{N}$ varies as $\mathrm{r}^{2}$.
MHT CET-2020
Moving Charges & Magnetism
153908
The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
1 $8 \times 10^{-24} \mathrm{Am}^{2}$
2 $15 \times 10^{-24} \mathrm{Am}^{2}$
3 $11 \times 10^{-24} \mathrm{Am}^{2}$
4 $6 \times 10^{-24} \mathrm{Am}^{2}$
Explanation:
A Given, $\mathrm{v}=2 \times 10^{6} \mathrm{~m} / \mathrm{s}, \mathrm{r}=0.5 \AA=0.5 \times 10^{-}$ ${ }^{10} \mathrm{~m}$ Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic moment $(\mathrm{M})=\frac{\mathrm{evr}}{2}$ $\mathrm{M}=\frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2}$ $\mathrm{M}=8 \times 10^{-24} \mathrm{Am}^{2}$
D Given, Radius of circular loop $(\mathrm{r})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Area of loop $(\mathrm{A})=\pi \mathrm{r}^{2}=\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}^{2}$ Current in the loop $(\mathrm{I})=0.1 \mathrm{~A}$ We know that, magnetic moment $(\mathrm{M})=\mathrm{I}$.A $\mathrm{M}=0.1 \times \pi\left(5 \times 10^{-2}\right)^{2}$ $\mathrm{M}=7.85 \times 10^{-4} \mathrm{~A}-\mathrm{m}^{2}$
CG PET- 2005
Moving Charges & Magnetism
153897
A magnet of magnetic moment $6 \mathrm{JT}^{-1}$ is aligned in the direction of magnetic field of $0.3 \mathrm{~T}$. The net work done to bring the magnet normal to the magnetic field is
1 $2 \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $1.8 \mathrm{~J}$
4 $2.4 \mathrm{~J}$
Explanation:
C Work required to turn the dipole is given by $\mathrm{W}=\mathrm{MB}\left(\cos \theta_{\mathrm{i}}-\cos \theta_{\mathrm{f}}\right)$ Here, $\theta_{i}=0^{\circ} \quad$ and $\theta_{\mathrm{f}}=90^{\circ}$ $\mathrm{W}=6 \times 0.3\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=1.8(1-0)$ $\mathrm{W}=1.8 \mathrm{~J}$
AP EAMCET-07.09.2021
Moving Charges & Magnetism
153903
Magnetic potential at any point on equatorial line of the magnetic dipole is
D Magnetic potential, $\mathrm{V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos \theta}{r^{2}}$ $\mathrm{~V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos 90^{\circ}}{\mathrm{r}^{2}}$ $\mathrm{~V}=0$
AP EAMCET (Medical)-07.10.2020
Moving Charges & Magnetism
153905
The magnetic moment of current (I) carrying circular coil of radius ( $r$ ) and number of turns (n) varies as:
1 $1 / \mathrm{r}^{2}$
2 $1 / \mathrm{r}$
3 $r$
4 $\mathrm{r}^{2}$
Explanation:
D We know that, the magnetic movement of a current carrying circular coil is given as- $\mathrm{M}=\mathrm{nIA}=\mathrm{nI} \pi \mathrm{r}^{2}$ Hence, magnetic moment of circular coil carrying current (I) of radius $r$ and number turns $\mathrm{N}$ varies as $\mathrm{r}^{2}$.
MHT CET-2020
Moving Charges & Magnetism
153908
The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
1 $8 \times 10^{-24} \mathrm{Am}^{2}$
2 $15 \times 10^{-24} \mathrm{Am}^{2}$
3 $11 \times 10^{-24} \mathrm{Am}^{2}$
4 $6 \times 10^{-24} \mathrm{Am}^{2}$
Explanation:
A Given, $\mathrm{v}=2 \times 10^{6} \mathrm{~m} / \mathrm{s}, \mathrm{r}=0.5 \AA=0.5 \times 10^{-}$ ${ }^{10} \mathrm{~m}$ Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic moment $(\mathrm{M})=\frac{\mathrm{evr}}{2}$ $\mathrm{M}=\frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2}$ $\mathrm{M}=8 \times 10^{-24} \mathrm{Am}^{2}$
D Given, Radius of circular loop $(\mathrm{r})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Area of loop $(\mathrm{A})=\pi \mathrm{r}^{2}=\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}^{2}$ Current in the loop $(\mathrm{I})=0.1 \mathrm{~A}$ We know that, magnetic moment $(\mathrm{M})=\mathrm{I}$.A $\mathrm{M}=0.1 \times \pi\left(5 \times 10^{-2}\right)^{2}$ $\mathrm{M}=7.85 \times 10^{-4} \mathrm{~A}-\mathrm{m}^{2}$
CG PET- 2005
Moving Charges & Magnetism
153897
A magnet of magnetic moment $6 \mathrm{JT}^{-1}$ is aligned in the direction of magnetic field of $0.3 \mathrm{~T}$. The net work done to bring the magnet normal to the magnetic field is
1 $2 \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $1.8 \mathrm{~J}$
4 $2.4 \mathrm{~J}$
Explanation:
C Work required to turn the dipole is given by $\mathrm{W}=\mathrm{MB}\left(\cos \theta_{\mathrm{i}}-\cos \theta_{\mathrm{f}}\right)$ Here, $\theta_{i}=0^{\circ} \quad$ and $\theta_{\mathrm{f}}=90^{\circ}$ $\mathrm{W}=6 \times 0.3\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=1.8(1-0)$ $\mathrm{W}=1.8 \mathrm{~J}$
AP EAMCET-07.09.2021
Moving Charges & Magnetism
153903
Magnetic potential at any point on equatorial line of the magnetic dipole is
D Magnetic potential, $\mathrm{V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos \theta}{r^{2}}$ $\mathrm{~V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos 90^{\circ}}{\mathrm{r}^{2}}$ $\mathrm{~V}=0$
AP EAMCET (Medical)-07.10.2020
Moving Charges & Magnetism
153905
The magnetic moment of current (I) carrying circular coil of radius ( $r$ ) and number of turns (n) varies as:
1 $1 / \mathrm{r}^{2}$
2 $1 / \mathrm{r}$
3 $r$
4 $\mathrm{r}^{2}$
Explanation:
D We know that, the magnetic movement of a current carrying circular coil is given as- $\mathrm{M}=\mathrm{nIA}=\mathrm{nI} \pi \mathrm{r}^{2}$ Hence, magnetic moment of circular coil carrying current (I) of radius $r$ and number turns $\mathrm{N}$ varies as $\mathrm{r}^{2}$.
MHT CET-2020
Moving Charges & Magnetism
153908
The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
1 $8 \times 10^{-24} \mathrm{Am}^{2}$
2 $15 \times 10^{-24} \mathrm{Am}^{2}$
3 $11 \times 10^{-24} \mathrm{Am}^{2}$
4 $6 \times 10^{-24} \mathrm{Am}^{2}$
Explanation:
A Given, $\mathrm{v}=2 \times 10^{6} \mathrm{~m} / \mathrm{s}, \mathrm{r}=0.5 \AA=0.5 \times 10^{-}$ ${ }^{10} \mathrm{~m}$ Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic moment $(\mathrm{M})=\frac{\mathrm{evr}}{2}$ $\mathrm{M}=\frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2}$ $\mathrm{M}=8 \times 10^{-24} \mathrm{Am}^{2}$
D Given, Radius of circular loop $(\mathrm{r})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Area of loop $(\mathrm{A})=\pi \mathrm{r}^{2}=\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}^{2}$ Current in the loop $(\mathrm{I})=0.1 \mathrm{~A}$ We know that, magnetic moment $(\mathrm{M})=\mathrm{I}$.A $\mathrm{M}=0.1 \times \pi\left(5 \times 10^{-2}\right)^{2}$ $\mathrm{M}=7.85 \times 10^{-4} \mathrm{~A}-\mathrm{m}^{2}$
CG PET- 2005
Moving Charges & Magnetism
153897
A magnet of magnetic moment $6 \mathrm{JT}^{-1}$ is aligned in the direction of magnetic field of $0.3 \mathrm{~T}$. The net work done to bring the magnet normal to the magnetic field is
1 $2 \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $1.8 \mathrm{~J}$
4 $2.4 \mathrm{~J}$
Explanation:
C Work required to turn the dipole is given by $\mathrm{W}=\mathrm{MB}\left(\cos \theta_{\mathrm{i}}-\cos \theta_{\mathrm{f}}\right)$ Here, $\theta_{i}=0^{\circ} \quad$ and $\theta_{\mathrm{f}}=90^{\circ}$ $\mathrm{W}=6 \times 0.3\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=1.8(1-0)$ $\mathrm{W}=1.8 \mathrm{~J}$
AP EAMCET-07.09.2021
Moving Charges & Magnetism
153903
Magnetic potential at any point on equatorial line of the magnetic dipole is
D Magnetic potential, $\mathrm{V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos \theta}{r^{2}}$ $\mathrm{~V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos 90^{\circ}}{\mathrm{r}^{2}}$ $\mathrm{~V}=0$
AP EAMCET (Medical)-07.10.2020
Moving Charges & Magnetism
153905
The magnetic moment of current (I) carrying circular coil of radius ( $r$ ) and number of turns (n) varies as:
1 $1 / \mathrm{r}^{2}$
2 $1 / \mathrm{r}$
3 $r$
4 $\mathrm{r}^{2}$
Explanation:
D We know that, the magnetic movement of a current carrying circular coil is given as- $\mathrm{M}=\mathrm{nIA}=\mathrm{nI} \pi \mathrm{r}^{2}$ Hence, magnetic moment of circular coil carrying current (I) of radius $r$ and number turns $\mathrm{N}$ varies as $\mathrm{r}^{2}$.
MHT CET-2020
Moving Charges & Magnetism
153908
The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
1 $8 \times 10^{-24} \mathrm{Am}^{2}$
2 $15 \times 10^{-24} \mathrm{Am}^{2}$
3 $11 \times 10^{-24} \mathrm{Am}^{2}$
4 $6 \times 10^{-24} \mathrm{Am}^{2}$
Explanation:
A Given, $\mathrm{v}=2 \times 10^{6} \mathrm{~m} / \mathrm{s}, \mathrm{r}=0.5 \AA=0.5 \times 10^{-}$ ${ }^{10} \mathrm{~m}$ Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic moment $(\mathrm{M})=\frac{\mathrm{evr}}{2}$ $\mathrm{M}=\frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2}$ $\mathrm{M}=8 \times 10^{-24} \mathrm{Am}^{2}$
D Given, Radius of circular loop $(\mathrm{r})=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ Area of loop $(\mathrm{A})=\pi \mathrm{r}^{2}=\pi\left(5 \times 10^{-2}\right)^{2} \mathrm{~m}^{2}$ Current in the loop $(\mathrm{I})=0.1 \mathrm{~A}$ We know that, magnetic moment $(\mathrm{M})=\mathrm{I}$.A $\mathrm{M}=0.1 \times \pi\left(5 \times 10^{-2}\right)^{2}$ $\mathrm{M}=7.85 \times 10^{-4} \mathrm{~A}-\mathrm{m}^{2}$
CG PET- 2005
Moving Charges & Magnetism
153897
A magnet of magnetic moment $6 \mathrm{JT}^{-1}$ is aligned in the direction of magnetic field of $0.3 \mathrm{~T}$. The net work done to bring the magnet normal to the magnetic field is
1 $2 \mathrm{~J}$
2 $1 \mathrm{~J}$
3 $1.8 \mathrm{~J}$
4 $2.4 \mathrm{~J}$
Explanation:
C Work required to turn the dipole is given by $\mathrm{W}=\mathrm{MB}\left(\cos \theta_{\mathrm{i}}-\cos \theta_{\mathrm{f}}\right)$ Here, $\theta_{i}=0^{\circ} \quad$ and $\theta_{\mathrm{f}}=90^{\circ}$ $\mathrm{W}=6 \times 0.3\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=1.8(1-0)$ $\mathrm{W}=1.8 \mathrm{~J}$
AP EAMCET-07.09.2021
Moving Charges & Magnetism
153903
Magnetic potential at any point on equatorial line of the magnetic dipole is
D Magnetic potential, $\mathrm{V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos \theta}{r^{2}}$ $\mathrm{~V}=\frac{\mu_{0}}{4 \pi} \times \frac{M \cos 90^{\circ}}{\mathrm{r}^{2}}$ $\mathrm{~V}=0$
AP EAMCET (Medical)-07.10.2020
Moving Charges & Magnetism
153905
The magnetic moment of current (I) carrying circular coil of radius ( $r$ ) and number of turns (n) varies as:
1 $1 / \mathrm{r}^{2}$
2 $1 / \mathrm{r}$
3 $r$
4 $\mathrm{r}^{2}$
Explanation:
D We know that, the magnetic movement of a current carrying circular coil is given as- $\mathrm{M}=\mathrm{nIA}=\mathrm{nI} \pi \mathrm{r}^{2}$ Hence, magnetic moment of circular coil carrying current (I) of radius $r$ and number turns $\mathrm{N}$ varies as $\mathrm{r}^{2}$.
MHT CET-2020
Moving Charges & Magnetism
153908
The electron in the hydrogen atom is moving with a speed of $2 \times 10^{6} \mathrm{~m} / \mathrm{s}$ in an orbit of radius $0.5 \AA$. The magnetic moment of the revolving electron is
1 $8 \times 10^{-24} \mathrm{Am}^{2}$
2 $15 \times 10^{-24} \mathrm{Am}^{2}$
3 $11 \times 10^{-24} \mathrm{Am}^{2}$
4 $6 \times 10^{-24} \mathrm{Am}^{2}$
Explanation:
A Given, $\mathrm{v}=2 \times 10^{6} \mathrm{~m} / \mathrm{s}, \mathrm{r}=0.5 \AA=0.5 \times 10^{-}$ ${ }^{10} \mathrm{~m}$ Charge of electron $(\mathrm{e})=1.6 \times 10^{-19} \mathrm{C}$ We know that, magnetic moment $(\mathrm{M})=\frac{\mathrm{evr}}{2}$ $\mathrm{M}=\frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2}$ $\mathrm{M}=8 \times 10^{-24} \mathrm{Am}^{2}$