NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153935
If the work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$, then the value of $n$ is
1 1
2 2
3 $\frac{1}{2}$
4 $\frac{1}{4}$
Explanation:
B Given, Angle of magnet, $\theta=90^{\circ}$ and $60^{\circ}$, Work done in turning the magnet through $90^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}(1-0)$ $\mathrm{W}_{1}=\mathrm{MB}$ Similarly, $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{2}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ Therefore $\mathrm{W}_{1}=\mathrm{nW}_{2}$ $\mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$ $\mathrm{n}=2$
VITEEE-2013
Moving Charges & Magnetism
153936
A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be
1 $\frac{4 \pi}{\mathrm{M}}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
3 $\sqrt{\frac{4 \pi \mathrm{i}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{\mathrm{i}}$
Explanation:
B Area of circular loop $\mathrm{A}=\pi \mathrm{r}^{2}$ $l=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{l}{2 \pi}$ Magnetic moment $(\mathrm{M})=\mathrm{i} . \mathrm{A}$ $\mathrm{M}=\mathrm{i} \pi \mathrm{r}^{2}$ $\mathrm{M}=\mathrm{i} . \pi \times \frac{l^{2}}{4 \pi^{2}}$ $l=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
VITEEE-2010
Moving Charges & Magnetism
153937
A circular coil of radius $10 \mathrm{~cm}$ and 100 turns carries a current $1 \mathrm{~A}$. What is the magnetic moment of the coil ?
1 $3.142 \times 10^{4} \mathrm{Am}^{2}$
2 $10^{4} \mathrm{Am}^{2}$
3 $3.142 \mathrm{Am}^{2}$
4 $3 \mathrm{Am}^{2}$
Explanation:
C Given that, Radius of circular coil $\mathrm{R}=10 \mathrm{~cm}$, Current $\mathrm{i}=1 \mathrm{~A}$ Number of turns $\mathrm{N}=100$ Magnetic moment $(\mathrm{M})=$ N.i.A $\mathrm{M}=100 \times 1 \times \pi \mathrm{R}^{2} \quad \quad\left[\because \mathrm{A}=\pi \mathrm{R}^{2}\right]$ $\mathrm{M}=100 \times \pi \times\left(10 \times 10^{-2}\right)^{2}$ $\mathrm{M}=100 \times \pi \times 100 \times 10^{-4}$ $\mathrm{M}=3.142 \mathrm{Am}^{2}$
Karnataka CET-2014
Moving Charges & Magnetism
153938
A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is $8.8 \times$ $10^{10} \mathrm{C} \mathrm{kg}^{-1}$. What is the mass of the electron? Given, charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$.
1 $1 \times 10^{-29} \mathrm{~kg}$
2 $0.1 \times 10^{-29} \mathrm{~kg}$
3 $1.1 \times 10^{-29} \mathrm{~kg}$
4 $1 / 11 \times 10^{-29} \mathrm{~kg}$
Explanation:
D Given that, Gyromagnetic ratio $\gamma=8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}$ We know that, Magnetic moment $\mu=\frac{\mathrm{evr}}{2}$ angular momentum $(\mathrm{L})=\mathrm{mvr}$ Gyromagnetic ratio is the ratio of magnetic moment of the particle to its angular momentum. $\therefore \quad \gamma=\frac{\mu}{\mathrm{L}}=\frac{\mathrm{evr} / 2}{\mathrm{mvr}}$ $\gamma=\frac{\mathrm{e}}{2 \mathrm{~m}}$ or $\quad 8.8 \times 10^{10}=\frac{1.6 \times 10^{-19}}{2 \mathrm{~m}}$ $\mathrm{m}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$ $\mathrm{m}=\frac{1}{11} \times 10^{-29} \mathrm{~kg}$
153935
If the work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$, then the value of $n$ is
1 1
2 2
3 $\frac{1}{2}$
4 $\frac{1}{4}$
Explanation:
B Given, Angle of magnet, $\theta=90^{\circ}$ and $60^{\circ}$, Work done in turning the magnet through $90^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}(1-0)$ $\mathrm{W}_{1}=\mathrm{MB}$ Similarly, $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{2}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ Therefore $\mathrm{W}_{1}=\mathrm{nW}_{2}$ $\mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$ $\mathrm{n}=2$
VITEEE-2013
Moving Charges & Magnetism
153936
A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be
1 $\frac{4 \pi}{\mathrm{M}}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
3 $\sqrt{\frac{4 \pi \mathrm{i}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{\mathrm{i}}$
Explanation:
B Area of circular loop $\mathrm{A}=\pi \mathrm{r}^{2}$ $l=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{l}{2 \pi}$ Magnetic moment $(\mathrm{M})=\mathrm{i} . \mathrm{A}$ $\mathrm{M}=\mathrm{i} \pi \mathrm{r}^{2}$ $\mathrm{M}=\mathrm{i} . \pi \times \frac{l^{2}}{4 \pi^{2}}$ $l=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
VITEEE-2010
Moving Charges & Magnetism
153937
A circular coil of radius $10 \mathrm{~cm}$ and 100 turns carries a current $1 \mathrm{~A}$. What is the magnetic moment of the coil ?
1 $3.142 \times 10^{4} \mathrm{Am}^{2}$
2 $10^{4} \mathrm{Am}^{2}$
3 $3.142 \mathrm{Am}^{2}$
4 $3 \mathrm{Am}^{2}$
Explanation:
C Given that, Radius of circular coil $\mathrm{R}=10 \mathrm{~cm}$, Current $\mathrm{i}=1 \mathrm{~A}$ Number of turns $\mathrm{N}=100$ Magnetic moment $(\mathrm{M})=$ N.i.A $\mathrm{M}=100 \times 1 \times \pi \mathrm{R}^{2} \quad \quad\left[\because \mathrm{A}=\pi \mathrm{R}^{2}\right]$ $\mathrm{M}=100 \times \pi \times\left(10 \times 10^{-2}\right)^{2}$ $\mathrm{M}=100 \times \pi \times 100 \times 10^{-4}$ $\mathrm{M}=3.142 \mathrm{Am}^{2}$
Karnataka CET-2014
Moving Charges & Magnetism
153938
A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is $8.8 \times$ $10^{10} \mathrm{C} \mathrm{kg}^{-1}$. What is the mass of the electron? Given, charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$.
1 $1 \times 10^{-29} \mathrm{~kg}$
2 $0.1 \times 10^{-29} \mathrm{~kg}$
3 $1.1 \times 10^{-29} \mathrm{~kg}$
4 $1 / 11 \times 10^{-29} \mathrm{~kg}$
Explanation:
D Given that, Gyromagnetic ratio $\gamma=8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}$ We know that, Magnetic moment $\mu=\frac{\mathrm{evr}}{2}$ angular momentum $(\mathrm{L})=\mathrm{mvr}$ Gyromagnetic ratio is the ratio of magnetic moment of the particle to its angular momentum. $\therefore \quad \gamma=\frac{\mu}{\mathrm{L}}=\frac{\mathrm{evr} / 2}{\mathrm{mvr}}$ $\gamma=\frac{\mathrm{e}}{2 \mathrm{~m}}$ or $\quad 8.8 \times 10^{10}=\frac{1.6 \times 10^{-19}}{2 \mathrm{~m}}$ $\mathrm{m}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$ $\mathrm{m}=\frac{1}{11} \times 10^{-29} \mathrm{~kg}$
153935
If the work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$, then the value of $n$ is
1 1
2 2
3 $\frac{1}{2}$
4 $\frac{1}{4}$
Explanation:
B Given, Angle of magnet, $\theta=90^{\circ}$ and $60^{\circ}$, Work done in turning the magnet through $90^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}(1-0)$ $\mathrm{W}_{1}=\mathrm{MB}$ Similarly, $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{2}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ Therefore $\mathrm{W}_{1}=\mathrm{nW}_{2}$ $\mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$ $\mathrm{n}=2$
VITEEE-2013
Moving Charges & Magnetism
153936
A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be
1 $\frac{4 \pi}{\mathrm{M}}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
3 $\sqrt{\frac{4 \pi \mathrm{i}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{\mathrm{i}}$
Explanation:
B Area of circular loop $\mathrm{A}=\pi \mathrm{r}^{2}$ $l=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{l}{2 \pi}$ Magnetic moment $(\mathrm{M})=\mathrm{i} . \mathrm{A}$ $\mathrm{M}=\mathrm{i} \pi \mathrm{r}^{2}$ $\mathrm{M}=\mathrm{i} . \pi \times \frac{l^{2}}{4 \pi^{2}}$ $l=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
VITEEE-2010
Moving Charges & Magnetism
153937
A circular coil of radius $10 \mathrm{~cm}$ and 100 turns carries a current $1 \mathrm{~A}$. What is the magnetic moment of the coil ?
1 $3.142 \times 10^{4} \mathrm{Am}^{2}$
2 $10^{4} \mathrm{Am}^{2}$
3 $3.142 \mathrm{Am}^{2}$
4 $3 \mathrm{Am}^{2}$
Explanation:
C Given that, Radius of circular coil $\mathrm{R}=10 \mathrm{~cm}$, Current $\mathrm{i}=1 \mathrm{~A}$ Number of turns $\mathrm{N}=100$ Magnetic moment $(\mathrm{M})=$ N.i.A $\mathrm{M}=100 \times 1 \times \pi \mathrm{R}^{2} \quad \quad\left[\because \mathrm{A}=\pi \mathrm{R}^{2}\right]$ $\mathrm{M}=100 \times \pi \times\left(10 \times 10^{-2}\right)^{2}$ $\mathrm{M}=100 \times \pi \times 100 \times 10^{-4}$ $\mathrm{M}=3.142 \mathrm{Am}^{2}$
Karnataka CET-2014
Moving Charges & Magnetism
153938
A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is $8.8 \times$ $10^{10} \mathrm{C} \mathrm{kg}^{-1}$. What is the mass of the electron? Given, charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$.
1 $1 \times 10^{-29} \mathrm{~kg}$
2 $0.1 \times 10^{-29} \mathrm{~kg}$
3 $1.1 \times 10^{-29} \mathrm{~kg}$
4 $1 / 11 \times 10^{-29} \mathrm{~kg}$
Explanation:
D Given that, Gyromagnetic ratio $\gamma=8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}$ We know that, Magnetic moment $\mu=\frac{\mathrm{evr}}{2}$ angular momentum $(\mathrm{L})=\mathrm{mvr}$ Gyromagnetic ratio is the ratio of magnetic moment of the particle to its angular momentum. $\therefore \quad \gamma=\frac{\mu}{\mathrm{L}}=\frac{\mathrm{evr} / 2}{\mathrm{mvr}}$ $\gamma=\frac{\mathrm{e}}{2 \mathrm{~m}}$ or $\quad 8.8 \times 10^{10}=\frac{1.6 \times 10^{-19}}{2 \mathrm{~m}}$ $\mathrm{m}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$ $\mathrm{m}=\frac{1}{11} \times 10^{-29} \mathrm{~kg}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153935
If the work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$, then the value of $n$ is
1 1
2 2
3 $\frac{1}{2}$
4 $\frac{1}{4}$
Explanation:
B Given, Angle of magnet, $\theta=90^{\circ}$ and $60^{\circ}$, Work done in turning the magnet through $90^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}(1-0)$ $\mathrm{W}_{1}=\mathrm{MB}$ Similarly, $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{2}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ Therefore $\mathrm{W}_{1}=\mathrm{nW}_{2}$ $\mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$ $\mathrm{n}=2$
VITEEE-2013
Moving Charges & Magnetism
153936
A straight wire carrying current $i$ is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be
1 $\frac{4 \pi}{\mathrm{M}}$
2 $\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
3 $\sqrt{\frac{4 \pi \mathrm{i}}{\mathrm{M}}}$
4 $\frac{\mathrm{M} \pi}{\mathrm{i}}$
Explanation:
B Area of circular loop $\mathrm{A}=\pi \mathrm{r}^{2}$ $l=2 \pi \mathrm{r}$ $\mathrm{r}=\frac{l}{2 \pi}$ Magnetic moment $(\mathrm{M})=\mathrm{i} . \mathrm{A}$ $\mathrm{M}=\mathrm{i} \pi \mathrm{r}^{2}$ $\mathrm{M}=\mathrm{i} . \pi \times \frac{l^{2}}{4 \pi^{2}}$ $l=\sqrt{\frac{4 \pi \mathrm{M}}{\mathrm{i}}}$
VITEEE-2010
Moving Charges & Magnetism
153937
A circular coil of radius $10 \mathrm{~cm}$ and 100 turns carries a current $1 \mathrm{~A}$. What is the magnetic moment of the coil ?
1 $3.142 \times 10^{4} \mathrm{Am}^{2}$
2 $10^{4} \mathrm{Am}^{2}$
3 $3.142 \mathrm{Am}^{2}$
4 $3 \mathrm{Am}^{2}$
Explanation:
C Given that, Radius of circular coil $\mathrm{R}=10 \mathrm{~cm}$, Current $\mathrm{i}=1 \mathrm{~A}$ Number of turns $\mathrm{N}=100$ Magnetic moment $(\mathrm{M})=$ N.i.A $\mathrm{M}=100 \times 1 \times \pi \mathrm{R}^{2} \quad \quad\left[\because \mathrm{A}=\pi \mathrm{R}^{2}\right]$ $\mathrm{M}=100 \times \pi \times\left(10 \times 10^{-2}\right)^{2}$ $\mathrm{M}=100 \times \pi \times 100 \times 10^{-4}$ $\mathrm{M}=3.142 \mathrm{Am}^{2}$
Karnataka CET-2014
Moving Charges & Magnetism
153938
A gyromagnetic ratio of the electron revolving in a circular orbit of hydrogen atom is $8.8 \times$ $10^{10} \mathrm{C} \mathrm{kg}^{-1}$. What is the mass of the electron? Given, charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$.
1 $1 \times 10^{-29} \mathrm{~kg}$
2 $0.1 \times 10^{-29} \mathrm{~kg}$
3 $1.1 \times 10^{-29} \mathrm{~kg}$
4 $1 / 11 \times 10^{-29} \mathrm{~kg}$
Explanation:
D Given that, Gyromagnetic ratio $\gamma=8.8 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}$ We know that, Magnetic moment $\mu=\frac{\mathrm{evr}}{2}$ angular momentum $(\mathrm{L})=\mathrm{mvr}$ Gyromagnetic ratio is the ratio of magnetic moment of the particle to its angular momentum. $\therefore \quad \gamma=\frac{\mu}{\mathrm{L}}=\frac{\mathrm{evr} / 2}{\mathrm{mvr}}$ $\gamma=\frac{\mathrm{e}}{2 \mathrm{~m}}$ or $\quad 8.8 \times 10^{10}=\frac{1.6 \times 10^{-19}}{2 \mathrm{~m}}$ $\mathrm{m}=\frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$ $\mathrm{m}=\frac{1}{11} \times 10^{-29} \mathrm{~kg}$
