154922
A wire of radius $R$ carries a current $I$. The magnetic induction inside the wire, at a distance $\mathbf{r}(\mathbf{r} \lt \mathbf{R})$, is
1 Independent of $\mathrm{R}$
2 Proportional to $\ln \mathrm{R}$
3 Proportional to $r$
4 Proportional to $1 / \mathrm{r}$
Explanation:
C According to question, We know, Ampere's circuital law $\text { f } \vec{B} \cdot d \vec{l}=\mu_{0} I_{\text {in }}$ And current enclosed by the closed path. $\mathrm{I}_{\mathrm{in}}=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}} \times \pi \mathrm{r}^{2}=\frac{\mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ Form equation (i) line integral of $\overrightarrow{\mathrm{B}}$ over this closed path is $\mathrm{B} \times 2 \pi \mathrm{r}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ $\mathrm{~B}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}}{2 \pi \mathrm{R}^{2}}$ $\mathrm{~B} \propto \mathrm{r}$ Hence, Magnetic field induction is proportional to $r$.
Assam CEE-2016
Electro Magnetic Induction
154923
In the given figure what will be the coefficient of mutual inductance
C $\therefore$ Magnetic indcution at a distance ' $r$ ' from wire $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}$ $\therefore$ Flux through the frame is obtained as $\phi_{12}=\int_{b}^{(a+b)} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{adr}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ia}}{2 \pi}[\ln \mathrm{n}]_{\mathrm{b}}^{\mathrm{a}+\mathrm{b}}$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)$ $\therefore \text { Mutual inductance }$ $\mathrm{M}_{12} =\frac{\phi_{12}}{\mathrm{I}}=\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi \mathrm{I}} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$ $\mathrm{M}_{12} =\frac{\mu_{\mathrm{o}} \mathrm{a}}{2 \pi} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$
JIPMER-2016
Electro Magnetic Induction
154924
The time constant coil is $3 \mathrm{~m}$ s. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
1 $54 \mathrm{mH}, 10 \Omega$
2 $14 \mathrm{mH}, 42 \Omega$
3 $42 \mathrm{mH}, 14 \Omega$
4 $14 \mathrm{mH}, 60 \Omega$
5 $54 \mathrm{mH}, 18 \Omega$
Explanation:
E Given that, Time constant $=3 \mathrm{~ms}=3 \times 10^{-3} \mathrm{~s}$ Resistance $(\mathrm{R})=90 \Omega$ Initially time constant $=\frac{\mathrm{L}}{\mathrm{r}}=3$ $\mathrm{L}=3 \mathrm{r}$ When resistance increase then new time constant, $\frac{\mathrm{L}}{\mathrm{r}+90}=0.5$ $\frac{3 \mathrm{r}}{\mathrm{r}+90}=0.5$ $3 \mathrm{r}-0.5 \mathrm{r}-45=0$ $2.5 \mathrm{r}=45$ $\mathrm{r}=18 \Omega$ Also, $\quad \mathrm{L}=3 \mathrm{r}$ $\mathrm{L}=3 \times 18$ $\mathrm{L}=54 \mathrm{mH}$
AP EAMCET -2010
Electro Magnetic Induction
154925
An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity $v$. Then, it travels
1 in a straight line without acceleration
2 with force in the direction of the field
3 in a circular path with a radius directly proportional to $\mathrm{v}^{2}$
4 in a circular path with a radius directly proportional to its velocity
Explanation:
D When particle enters perpendicular to uniform magnetic field, then it travels in a circular path. Then, $\quad \mathrm{f}_{\text {centripetal }}=\mathrm{f}_{\text {magnetic }}$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$ Where, $\mathrm{v} \stackrel{\mathrm{r}}{=}$ Velocity of particle $\mathrm{B}=$ Magnetic field $\mathrm{m}=$ Mass of particle $\mathrm{r}=$ Radius of path on which particle moves Hence, $r=\frac{m v}{B q}, \quad r \propto v$
AP EAMCET -2014
Electro Magnetic Induction
154926
A square loop of wire, side length $10 \mathrm{~cm}$ is placed at angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \mathrm{~T}$ to zero in 0.7 seconds. The induced current in the loop (its resistance is $1 \Omega$ ) is
154922
A wire of radius $R$ carries a current $I$. The magnetic induction inside the wire, at a distance $\mathbf{r}(\mathbf{r} \lt \mathbf{R})$, is
1 Independent of $\mathrm{R}$
2 Proportional to $\ln \mathrm{R}$
3 Proportional to $r$
4 Proportional to $1 / \mathrm{r}$
Explanation:
C According to question, We know, Ampere's circuital law $\text { f } \vec{B} \cdot d \vec{l}=\mu_{0} I_{\text {in }}$ And current enclosed by the closed path. $\mathrm{I}_{\mathrm{in}}=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}} \times \pi \mathrm{r}^{2}=\frac{\mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ Form equation (i) line integral of $\overrightarrow{\mathrm{B}}$ over this closed path is $\mathrm{B} \times 2 \pi \mathrm{r}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ $\mathrm{~B}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}}{2 \pi \mathrm{R}^{2}}$ $\mathrm{~B} \propto \mathrm{r}$ Hence, Magnetic field induction is proportional to $r$.
Assam CEE-2016
Electro Magnetic Induction
154923
In the given figure what will be the coefficient of mutual inductance
C $\therefore$ Magnetic indcution at a distance ' $r$ ' from wire $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}$ $\therefore$ Flux through the frame is obtained as $\phi_{12}=\int_{b}^{(a+b)} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{adr}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ia}}{2 \pi}[\ln \mathrm{n}]_{\mathrm{b}}^{\mathrm{a}+\mathrm{b}}$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)$ $\therefore \text { Mutual inductance }$ $\mathrm{M}_{12} =\frac{\phi_{12}}{\mathrm{I}}=\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi \mathrm{I}} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$ $\mathrm{M}_{12} =\frac{\mu_{\mathrm{o}} \mathrm{a}}{2 \pi} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$
JIPMER-2016
Electro Magnetic Induction
154924
The time constant coil is $3 \mathrm{~m}$ s. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
1 $54 \mathrm{mH}, 10 \Omega$
2 $14 \mathrm{mH}, 42 \Omega$
3 $42 \mathrm{mH}, 14 \Omega$
4 $14 \mathrm{mH}, 60 \Omega$
5 $54 \mathrm{mH}, 18 \Omega$
Explanation:
E Given that, Time constant $=3 \mathrm{~ms}=3 \times 10^{-3} \mathrm{~s}$ Resistance $(\mathrm{R})=90 \Omega$ Initially time constant $=\frac{\mathrm{L}}{\mathrm{r}}=3$ $\mathrm{L}=3 \mathrm{r}$ When resistance increase then new time constant, $\frac{\mathrm{L}}{\mathrm{r}+90}=0.5$ $\frac{3 \mathrm{r}}{\mathrm{r}+90}=0.5$ $3 \mathrm{r}-0.5 \mathrm{r}-45=0$ $2.5 \mathrm{r}=45$ $\mathrm{r}=18 \Omega$ Also, $\quad \mathrm{L}=3 \mathrm{r}$ $\mathrm{L}=3 \times 18$ $\mathrm{L}=54 \mathrm{mH}$
AP EAMCET -2010
Electro Magnetic Induction
154925
An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity $v$. Then, it travels
1 in a straight line without acceleration
2 with force in the direction of the field
3 in a circular path with a radius directly proportional to $\mathrm{v}^{2}$
4 in a circular path with a radius directly proportional to its velocity
Explanation:
D When particle enters perpendicular to uniform magnetic field, then it travels in a circular path. Then, $\quad \mathrm{f}_{\text {centripetal }}=\mathrm{f}_{\text {magnetic }}$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$ Where, $\mathrm{v} \stackrel{\mathrm{r}}{=}$ Velocity of particle $\mathrm{B}=$ Magnetic field $\mathrm{m}=$ Mass of particle $\mathrm{r}=$ Radius of path on which particle moves Hence, $r=\frac{m v}{B q}, \quad r \propto v$
AP EAMCET -2014
Electro Magnetic Induction
154926
A square loop of wire, side length $10 \mathrm{~cm}$ is placed at angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \mathrm{~T}$ to zero in 0.7 seconds. The induced current in the loop (its resistance is $1 \Omega$ ) is
154922
A wire of radius $R$ carries a current $I$. The magnetic induction inside the wire, at a distance $\mathbf{r}(\mathbf{r} \lt \mathbf{R})$, is
1 Independent of $\mathrm{R}$
2 Proportional to $\ln \mathrm{R}$
3 Proportional to $r$
4 Proportional to $1 / \mathrm{r}$
Explanation:
C According to question, We know, Ampere's circuital law $\text { f } \vec{B} \cdot d \vec{l}=\mu_{0} I_{\text {in }}$ And current enclosed by the closed path. $\mathrm{I}_{\mathrm{in}}=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}} \times \pi \mathrm{r}^{2}=\frac{\mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ Form equation (i) line integral of $\overrightarrow{\mathrm{B}}$ over this closed path is $\mathrm{B} \times 2 \pi \mathrm{r}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ $\mathrm{~B}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}}{2 \pi \mathrm{R}^{2}}$ $\mathrm{~B} \propto \mathrm{r}$ Hence, Magnetic field induction is proportional to $r$.
Assam CEE-2016
Electro Magnetic Induction
154923
In the given figure what will be the coefficient of mutual inductance
C $\therefore$ Magnetic indcution at a distance ' $r$ ' from wire $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}$ $\therefore$ Flux through the frame is obtained as $\phi_{12}=\int_{b}^{(a+b)} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{adr}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ia}}{2 \pi}[\ln \mathrm{n}]_{\mathrm{b}}^{\mathrm{a}+\mathrm{b}}$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)$ $\therefore \text { Mutual inductance }$ $\mathrm{M}_{12} =\frac{\phi_{12}}{\mathrm{I}}=\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi \mathrm{I}} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$ $\mathrm{M}_{12} =\frac{\mu_{\mathrm{o}} \mathrm{a}}{2 \pi} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$
JIPMER-2016
Electro Magnetic Induction
154924
The time constant coil is $3 \mathrm{~m}$ s. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
1 $54 \mathrm{mH}, 10 \Omega$
2 $14 \mathrm{mH}, 42 \Omega$
3 $42 \mathrm{mH}, 14 \Omega$
4 $14 \mathrm{mH}, 60 \Omega$
5 $54 \mathrm{mH}, 18 \Omega$
Explanation:
E Given that, Time constant $=3 \mathrm{~ms}=3 \times 10^{-3} \mathrm{~s}$ Resistance $(\mathrm{R})=90 \Omega$ Initially time constant $=\frac{\mathrm{L}}{\mathrm{r}}=3$ $\mathrm{L}=3 \mathrm{r}$ When resistance increase then new time constant, $\frac{\mathrm{L}}{\mathrm{r}+90}=0.5$ $\frac{3 \mathrm{r}}{\mathrm{r}+90}=0.5$ $3 \mathrm{r}-0.5 \mathrm{r}-45=0$ $2.5 \mathrm{r}=45$ $\mathrm{r}=18 \Omega$ Also, $\quad \mathrm{L}=3 \mathrm{r}$ $\mathrm{L}=3 \times 18$ $\mathrm{L}=54 \mathrm{mH}$
AP EAMCET -2010
Electro Magnetic Induction
154925
An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity $v$. Then, it travels
1 in a straight line without acceleration
2 with force in the direction of the field
3 in a circular path with a radius directly proportional to $\mathrm{v}^{2}$
4 in a circular path with a radius directly proportional to its velocity
Explanation:
D When particle enters perpendicular to uniform magnetic field, then it travels in a circular path. Then, $\quad \mathrm{f}_{\text {centripetal }}=\mathrm{f}_{\text {magnetic }}$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$ Where, $\mathrm{v} \stackrel{\mathrm{r}}{=}$ Velocity of particle $\mathrm{B}=$ Magnetic field $\mathrm{m}=$ Mass of particle $\mathrm{r}=$ Radius of path on which particle moves Hence, $r=\frac{m v}{B q}, \quad r \propto v$
AP EAMCET -2014
Electro Magnetic Induction
154926
A square loop of wire, side length $10 \mathrm{~cm}$ is placed at angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \mathrm{~T}$ to zero in 0.7 seconds. The induced current in the loop (its resistance is $1 \Omega$ ) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154922
A wire of radius $R$ carries a current $I$. The magnetic induction inside the wire, at a distance $\mathbf{r}(\mathbf{r} \lt \mathbf{R})$, is
1 Independent of $\mathrm{R}$
2 Proportional to $\ln \mathrm{R}$
3 Proportional to $r$
4 Proportional to $1 / \mathrm{r}$
Explanation:
C According to question, We know, Ampere's circuital law $\text { f } \vec{B} \cdot d \vec{l}=\mu_{0} I_{\text {in }}$ And current enclosed by the closed path. $\mathrm{I}_{\mathrm{in}}=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}} \times \pi \mathrm{r}^{2}=\frac{\mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ Form equation (i) line integral of $\overrightarrow{\mathrm{B}}$ over this closed path is $\mathrm{B} \times 2 \pi \mathrm{r}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ $\mathrm{~B}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}}{2 \pi \mathrm{R}^{2}}$ $\mathrm{~B} \propto \mathrm{r}$ Hence, Magnetic field induction is proportional to $r$.
Assam CEE-2016
Electro Magnetic Induction
154923
In the given figure what will be the coefficient of mutual inductance
C $\therefore$ Magnetic indcution at a distance ' $r$ ' from wire $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}$ $\therefore$ Flux through the frame is obtained as $\phi_{12}=\int_{b}^{(a+b)} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{adr}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ia}}{2 \pi}[\ln \mathrm{n}]_{\mathrm{b}}^{\mathrm{a}+\mathrm{b}}$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)$ $\therefore \text { Mutual inductance }$ $\mathrm{M}_{12} =\frac{\phi_{12}}{\mathrm{I}}=\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi \mathrm{I}} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$ $\mathrm{M}_{12} =\frac{\mu_{\mathrm{o}} \mathrm{a}}{2 \pi} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$
JIPMER-2016
Electro Magnetic Induction
154924
The time constant coil is $3 \mathrm{~m}$ s. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
1 $54 \mathrm{mH}, 10 \Omega$
2 $14 \mathrm{mH}, 42 \Omega$
3 $42 \mathrm{mH}, 14 \Omega$
4 $14 \mathrm{mH}, 60 \Omega$
5 $54 \mathrm{mH}, 18 \Omega$
Explanation:
E Given that, Time constant $=3 \mathrm{~ms}=3 \times 10^{-3} \mathrm{~s}$ Resistance $(\mathrm{R})=90 \Omega$ Initially time constant $=\frac{\mathrm{L}}{\mathrm{r}}=3$ $\mathrm{L}=3 \mathrm{r}$ When resistance increase then new time constant, $\frac{\mathrm{L}}{\mathrm{r}+90}=0.5$ $\frac{3 \mathrm{r}}{\mathrm{r}+90}=0.5$ $3 \mathrm{r}-0.5 \mathrm{r}-45=0$ $2.5 \mathrm{r}=45$ $\mathrm{r}=18 \Omega$ Also, $\quad \mathrm{L}=3 \mathrm{r}$ $\mathrm{L}=3 \times 18$ $\mathrm{L}=54 \mathrm{mH}$
AP EAMCET -2010
Electro Magnetic Induction
154925
An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity $v$. Then, it travels
1 in a straight line without acceleration
2 with force in the direction of the field
3 in a circular path with a radius directly proportional to $\mathrm{v}^{2}$
4 in a circular path with a radius directly proportional to its velocity
Explanation:
D When particle enters perpendicular to uniform magnetic field, then it travels in a circular path. Then, $\quad \mathrm{f}_{\text {centripetal }}=\mathrm{f}_{\text {magnetic }}$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$ Where, $\mathrm{v} \stackrel{\mathrm{r}}{=}$ Velocity of particle $\mathrm{B}=$ Magnetic field $\mathrm{m}=$ Mass of particle $\mathrm{r}=$ Radius of path on which particle moves Hence, $r=\frac{m v}{B q}, \quad r \propto v$
AP EAMCET -2014
Electro Magnetic Induction
154926
A square loop of wire, side length $10 \mathrm{~cm}$ is placed at angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \mathrm{~T}$ to zero in 0.7 seconds. The induced current in the loop (its resistance is $1 \Omega$ ) is
154922
A wire of radius $R$ carries a current $I$. The magnetic induction inside the wire, at a distance $\mathbf{r}(\mathbf{r} \lt \mathbf{R})$, is
1 Independent of $\mathrm{R}$
2 Proportional to $\ln \mathrm{R}$
3 Proportional to $r$
4 Proportional to $1 / \mathrm{r}$
Explanation:
C According to question, We know, Ampere's circuital law $\text { f } \vec{B} \cdot d \vec{l}=\mu_{0} I_{\text {in }}$ And current enclosed by the closed path. $\mathrm{I}_{\mathrm{in}}=\frac{\mathrm{I}}{\pi \mathrm{R}^{2}} \times \pi \mathrm{r}^{2}=\frac{\mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ Form equation (i) line integral of $\overrightarrow{\mathrm{B}}$ over this closed path is $\mathrm{B} \times 2 \pi \mathrm{r}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}^{2}}{\mathrm{R}^{2}}$ $\mathrm{~B}=\frac{\mu_{\mathrm{o}} \mathrm{Ir}}{2 \pi \mathrm{R}^{2}}$ $\mathrm{~B} \propto \mathrm{r}$ Hence, Magnetic field induction is proportional to $r$.
Assam CEE-2016
Electro Magnetic Induction
154923
In the given figure what will be the coefficient of mutual inductance
C $\therefore$ Magnetic indcution at a distance ' $r$ ' from wire $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{r}}$ $\therefore$ Flux through the frame is obtained as $\phi_{12}=\int_{b}^{(a+b)} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{adr}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ia}}{2 \pi}[\ln \mathrm{n}]_{\mathrm{b}}^{\mathrm{a}+\mathrm{b}}$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)$ $\phi_{12} =\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi} \ln \left(1+\frac{\mathrm{a}}{\mathrm{b}}\right)$ $\therefore \text { Mutual inductance }$ $\mathrm{M}_{12} =\frac{\phi_{12}}{\mathrm{I}}=\frac{\mu_{\mathrm{o}} \mathrm{Ia}}{2 \pi \mathrm{I}} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$ $\mathrm{M}_{12} =\frac{\mu_{\mathrm{o}} \mathrm{a}}{2 \pi} \ln \left[1+\frac{\mathrm{a}}{\mathrm{b}}\right]$
JIPMER-2016
Electro Magnetic Induction
154924
The time constant coil is $3 \mathrm{~m}$ s. When a $90 \Omega$ resistance is joined in series, then the time constant becomes $0.5 \mathrm{~m} \mathrm{~s}$. The inductance and the resistance of the coil are
1 $54 \mathrm{mH}, 10 \Omega$
2 $14 \mathrm{mH}, 42 \Omega$
3 $42 \mathrm{mH}, 14 \Omega$
4 $14 \mathrm{mH}, 60 \Omega$
5 $54 \mathrm{mH}, 18 \Omega$
Explanation:
E Given that, Time constant $=3 \mathrm{~ms}=3 \times 10^{-3} \mathrm{~s}$ Resistance $(\mathrm{R})=90 \Omega$ Initially time constant $=\frac{\mathrm{L}}{\mathrm{r}}=3$ $\mathrm{L}=3 \mathrm{r}$ When resistance increase then new time constant, $\frac{\mathrm{L}}{\mathrm{r}+90}=0.5$ $\frac{3 \mathrm{r}}{\mathrm{r}+90}=0.5$ $3 \mathrm{r}-0.5 \mathrm{r}-45=0$ $2.5 \mathrm{r}=45$ $\mathrm{r}=18 \Omega$ Also, $\quad \mathrm{L}=3 \mathrm{r}$ $\mathrm{L}=3 \times 18$ $\mathrm{L}=54 \mathrm{mH}$
AP EAMCET -2010
Electro Magnetic Induction
154925
An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity $v$. Then, it travels
1 in a straight line without acceleration
2 with force in the direction of the field
3 in a circular path with a radius directly proportional to $\mathrm{v}^{2}$
4 in a circular path with a radius directly proportional to its velocity
Explanation:
D When particle enters perpendicular to uniform magnetic field, then it travels in a circular path. Then, $\quad \mathrm{f}_{\text {centripetal }}=\mathrm{f}_{\text {magnetic }}$ $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$ Where, $\mathrm{v} \stackrel{\mathrm{r}}{=}$ Velocity of particle $\mathrm{B}=$ Magnetic field $\mathrm{m}=$ Mass of particle $\mathrm{r}=$ Radius of path on which particle moves Hence, $r=\frac{m v}{B q}, \quad r \propto v$
AP EAMCET -2014
Electro Magnetic Induction
154926
A square loop of wire, side length $10 \mathrm{~cm}$ is placed at angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \mathrm{~T}$ to zero in 0.7 seconds. The induced current in the loop (its resistance is $1 \Omega$ ) is
