153915
If the dipole moment of a short bar magnet is 1.25 $A-m^{2}$, find the magnetic field on its axis at a distance of $0.5 \mathrm{~m}$ from the centre of the magnet.
C Given that, $\mathrm{M}=1.25 \mathrm{Am}^{2}$ $\mathrm{r}=0.5 \mathrm{~m}$ Then, magnetic field $\mathrm{B} =\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^{3}}$ $\mathrm{~B} =\frac{10^{-7} \times 2 \times 1.25}{(0.5)^{3}}$ $\mathrm{~B} =2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}$
AP EAMCET (22.09.2020) Shift-I
Moving Charges & Magnetism
153916
An electron having charge ' $e$ ' is moving with a constant speed $v$ along a circle of radius $r$. Its magnetic moment will be
1 e $v \mathrm{r}$
2 $\frac{\mathrm{e} v \mathrm{r}}{2}$
3 $2 \pi \mathrm{rev}$
4 Zero
Explanation:
B Magnetic moment $\mathrm{q}=\mathrm{it}$ $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$ $\mathrm{i}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ $\mathrm{m}=\mathrm{i} \times \mathrm{A}$ $\mathrm{m}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{~m}=\frac{\mathrm{evr}}{2}$
UPSEE 2019
Moving Charges & Magnetism
153917
If a circular coil of radius $3 \mathrm{~cm}$ having 10 turns carries a current $0.2 \mathrm{~A}$, then magnetic moment of the coil is
1 $5.65 \times 10^{-3} \mathrm{Am}^{2}$
2 $6.56 \times 10^{-3} \mathrm{Am}^{2}$
3 $4 \times 10^{-3} \mathrm{Am}^{2}$
4 $3.5 \times 10^{-3} \mathrm{Am}^{2}$
Explanation:
A Given, radius of coil $=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ Number of turns $(\mathrm{N})=10$, current $(\mathrm{I})=0.2 \mathrm{~A}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=3.14 \times\left(3 \times 10^{-2}\right)^{2}$ We know that, magnetic moment $(\mathrm{M})=$ NIA $\mathrm{M}=10 \times 0.2 \times 3.14 \times\left(3 \times 10^{-2}\right)^{2}$ $\mathrm{M}=0.00565 \mathrm{Am}^{2}$ $\mathrm{M}=5.65 \times 10^{-3} \mathrm{Am}^{2}$
MHT-CET 2019
Moving Charges & Magnetism
153919
An electron having charge ' $e$ ' revolves around the nucleus in a circular orbit of radius ' $r$ ' at a speed of ' $v$ '. Its magnetic moment is given by
1 $\frac{\text { erv }}{2}$
2 $\operatorname{er}^{2} v$
3 erv
4 $\frac{\text { erv }}{2 \pi}$
Explanation:
A Given, charge (q) $=$ e, orbit radius $=r$, speed of electron $=\mathrm{v}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{e}}{\mathrm{t}}$ $\mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$ $\quad \mathrm{I}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ Putting the value of I and A in equation (i), we get- $\mathrm{M}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{evr}}{2}$
153915
If the dipole moment of a short bar magnet is 1.25 $A-m^{2}$, find the magnetic field on its axis at a distance of $0.5 \mathrm{~m}$ from the centre of the magnet.
C Given that, $\mathrm{M}=1.25 \mathrm{Am}^{2}$ $\mathrm{r}=0.5 \mathrm{~m}$ Then, magnetic field $\mathrm{B} =\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^{3}}$ $\mathrm{~B} =\frac{10^{-7} \times 2 \times 1.25}{(0.5)^{3}}$ $\mathrm{~B} =2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}$
AP EAMCET (22.09.2020) Shift-I
Moving Charges & Magnetism
153916
An electron having charge ' $e$ ' is moving with a constant speed $v$ along a circle of radius $r$. Its magnetic moment will be
1 e $v \mathrm{r}$
2 $\frac{\mathrm{e} v \mathrm{r}}{2}$
3 $2 \pi \mathrm{rev}$
4 Zero
Explanation:
B Magnetic moment $\mathrm{q}=\mathrm{it}$ $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$ $\mathrm{i}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ $\mathrm{m}=\mathrm{i} \times \mathrm{A}$ $\mathrm{m}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{~m}=\frac{\mathrm{evr}}{2}$
UPSEE 2019
Moving Charges & Magnetism
153917
If a circular coil of radius $3 \mathrm{~cm}$ having 10 turns carries a current $0.2 \mathrm{~A}$, then magnetic moment of the coil is
1 $5.65 \times 10^{-3} \mathrm{Am}^{2}$
2 $6.56 \times 10^{-3} \mathrm{Am}^{2}$
3 $4 \times 10^{-3} \mathrm{Am}^{2}$
4 $3.5 \times 10^{-3} \mathrm{Am}^{2}$
Explanation:
A Given, radius of coil $=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ Number of turns $(\mathrm{N})=10$, current $(\mathrm{I})=0.2 \mathrm{~A}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=3.14 \times\left(3 \times 10^{-2}\right)^{2}$ We know that, magnetic moment $(\mathrm{M})=$ NIA $\mathrm{M}=10 \times 0.2 \times 3.14 \times\left(3 \times 10^{-2}\right)^{2}$ $\mathrm{M}=0.00565 \mathrm{Am}^{2}$ $\mathrm{M}=5.65 \times 10^{-3} \mathrm{Am}^{2}$
MHT-CET 2019
Moving Charges & Magnetism
153919
An electron having charge ' $e$ ' revolves around the nucleus in a circular orbit of radius ' $r$ ' at a speed of ' $v$ '. Its magnetic moment is given by
1 $\frac{\text { erv }}{2}$
2 $\operatorname{er}^{2} v$
3 erv
4 $\frac{\text { erv }}{2 \pi}$
Explanation:
A Given, charge (q) $=$ e, orbit radius $=r$, speed of electron $=\mathrm{v}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{e}}{\mathrm{t}}$ $\mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$ $\quad \mathrm{I}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ Putting the value of I and A in equation (i), we get- $\mathrm{M}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{evr}}{2}$
153915
If the dipole moment of a short bar magnet is 1.25 $A-m^{2}$, find the magnetic field on its axis at a distance of $0.5 \mathrm{~m}$ from the centre of the magnet.
C Given that, $\mathrm{M}=1.25 \mathrm{Am}^{2}$ $\mathrm{r}=0.5 \mathrm{~m}$ Then, magnetic field $\mathrm{B} =\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^{3}}$ $\mathrm{~B} =\frac{10^{-7} \times 2 \times 1.25}{(0.5)^{3}}$ $\mathrm{~B} =2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}$
AP EAMCET (22.09.2020) Shift-I
Moving Charges & Magnetism
153916
An electron having charge ' $e$ ' is moving with a constant speed $v$ along a circle of radius $r$. Its magnetic moment will be
1 e $v \mathrm{r}$
2 $\frac{\mathrm{e} v \mathrm{r}}{2}$
3 $2 \pi \mathrm{rev}$
4 Zero
Explanation:
B Magnetic moment $\mathrm{q}=\mathrm{it}$ $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$ $\mathrm{i}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ $\mathrm{m}=\mathrm{i} \times \mathrm{A}$ $\mathrm{m}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{~m}=\frac{\mathrm{evr}}{2}$
UPSEE 2019
Moving Charges & Magnetism
153917
If a circular coil of radius $3 \mathrm{~cm}$ having 10 turns carries a current $0.2 \mathrm{~A}$, then magnetic moment of the coil is
1 $5.65 \times 10^{-3} \mathrm{Am}^{2}$
2 $6.56 \times 10^{-3} \mathrm{Am}^{2}$
3 $4 \times 10^{-3} \mathrm{Am}^{2}$
4 $3.5 \times 10^{-3} \mathrm{Am}^{2}$
Explanation:
A Given, radius of coil $=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ Number of turns $(\mathrm{N})=10$, current $(\mathrm{I})=0.2 \mathrm{~A}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=3.14 \times\left(3 \times 10^{-2}\right)^{2}$ We know that, magnetic moment $(\mathrm{M})=$ NIA $\mathrm{M}=10 \times 0.2 \times 3.14 \times\left(3 \times 10^{-2}\right)^{2}$ $\mathrm{M}=0.00565 \mathrm{Am}^{2}$ $\mathrm{M}=5.65 \times 10^{-3} \mathrm{Am}^{2}$
MHT-CET 2019
Moving Charges & Magnetism
153919
An electron having charge ' $e$ ' revolves around the nucleus in a circular orbit of radius ' $r$ ' at a speed of ' $v$ '. Its magnetic moment is given by
1 $\frac{\text { erv }}{2}$
2 $\operatorname{er}^{2} v$
3 erv
4 $\frac{\text { erv }}{2 \pi}$
Explanation:
A Given, charge (q) $=$ e, orbit radius $=r$, speed of electron $=\mathrm{v}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{e}}{\mathrm{t}}$ $\mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$ $\quad \mathrm{I}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ Putting the value of I and A in equation (i), we get- $\mathrm{M}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{evr}}{2}$
153915
If the dipole moment of a short bar magnet is 1.25 $A-m^{2}$, find the magnetic field on its axis at a distance of $0.5 \mathrm{~m}$ from the centre of the magnet.
C Given that, $\mathrm{M}=1.25 \mathrm{Am}^{2}$ $\mathrm{r}=0.5 \mathrm{~m}$ Then, magnetic field $\mathrm{B} =\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{r}^{3}}$ $\mathrm{~B} =\frac{10^{-7} \times 2 \times 1.25}{(0.5)^{3}}$ $\mathrm{~B} =2 \times 10^{-6} \mathrm{NA}^{-1} \mathrm{~m}^{-1}$
AP EAMCET (22.09.2020) Shift-I
Moving Charges & Magnetism
153916
An electron having charge ' $e$ ' is moving with a constant speed $v$ along a circle of radius $r$. Its magnetic moment will be
1 e $v \mathrm{r}$
2 $\frac{\mathrm{e} v \mathrm{r}}{2}$
3 $2 \pi \mathrm{rev}$
4 Zero
Explanation:
B Magnetic moment $\mathrm{q}=\mathrm{it}$ $\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$ $\mathrm{i}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}}$ $\mathrm{m}=\mathrm{i} \times \mathrm{A}$ $\mathrm{m}=\frac{\mathrm{qv}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{~m}=\frac{\mathrm{evr}}{2}$
UPSEE 2019
Moving Charges & Magnetism
153917
If a circular coil of radius $3 \mathrm{~cm}$ having 10 turns carries a current $0.2 \mathrm{~A}$, then magnetic moment of the coil is
1 $5.65 \times 10^{-3} \mathrm{Am}^{2}$
2 $6.56 \times 10^{-3} \mathrm{Am}^{2}$
3 $4 \times 10^{-3} \mathrm{Am}^{2}$
4 $3.5 \times 10^{-3} \mathrm{Am}^{2}$
Explanation:
A Given, radius of coil $=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ Number of turns $(\mathrm{N})=10$, current $(\mathrm{I})=0.2 \mathrm{~A}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ $\mathrm{A}=3.14 \times\left(3 \times 10^{-2}\right)^{2}$ We know that, magnetic moment $(\mathrm{M})=$ NIA $\mathrm{M}=10 \times 0.2 \times 3.14 \times\left(3 \times 10^{-2}\right)^{2}$ $\mathrm{M}=0.00565 \mathrm{Am}^{2}$ $\mathrm{M}=5.65 \times 10^{-3} \mathrm{Am}^{2}$
MHT-CET 2019
Moving Charges & Magnetism
153919
An electron having charge ' $e$ ' revolves around the nucleus in a circular orbit of radius ' $r$ ' at a speed of ' $v$ '. Its magnetic moment is given by
1 $\frac{\text { erv }}{2}$
2 $\operatorname{er}^{2} v$
3 erv
4 $\frac{\text { erv }}{2 \pi}$
Explanation:
A Given, charge (q) $=$ e, orbit radius $=r$, speed of electron $=\mathrm{v}$ Magnetic moment $(\mathrm{M})=\mathrm{IA}$ $\mathrm{I}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{\mathrm{e}}{\mathrm{t}}$ $\mathrm{t}=\frac{2 \pi \mathrm{r}}{\mathrm{v}}$ $\quad \mathrm{I}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{v}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$ Area of circle $(\mathrm{A})=\pi \mathrm{r}^{2}$ Putting the value of I and A in equation (i), we get- $\mathrm{M}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\frac{\mathrm{evr}}{2}$
