154746
A long straight wire carrying current $I$ and a rectangular frame with side lengths a and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is
1 $\frac{\mu_{0}}{2 \pi} \mathrm{ab}$
2 $\frac{\mu}{4 \pi} \frac{b}{a}$
3 $\frac{\mu_{0} \mathrm{a}}{2 \pi} \ln 2$
4 $\frac{\mu_{0} b}{2 \pi} \ln 2$
Explanation:
D Magnetic field at a distance $r$ from wire $B=\frac{\mu_{0} I}{2 \pi r}$ Flux through the frame is obtained by $\phi_{12}=\int_{\mathrm{a}}^{(\mathrm{a}+\mathrm{a})} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{bdr}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{2 \mathrm{a}}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln \frac{2 \mathrm{a}}{\mathrm{a}}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2$ Mutual induction $\mathrm{M}_{12}=\frac{\phi_{12}}{\mathrm{I}}=\frac{\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2}{\mathrm{I}}$ $\mathrm{M}_{12}=\frac{\mu_{0} \mathrm{~b}}{2 \pi} \ln 2$
TS EAMCET 06.08.2021
Electro Magnetic Induction
154747
A pair of adjacent coils has a mutual inductance of $1.5 \mathrm{H}$. If the current in one coil changes from 0 to $20 \mathrm{~A}$ in 0.5 sec. what is the change of flux linkage with the other coil?
154749
A coil of inductance $L$ is divided into four equal parts and all the parts are connected in parallel. The effective inductance of the combination is
1 $\frac{L}{4}$
2 $\frac{L}{8}$
3 $\frac{L}{16}$
4 $4 L$
Explanation:
C Given, inductance $=\mathrm{L}$ Inductance of each part $=\frac{\mathrm{L}}{4}$ When inductance connected in parallel- $\frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{4}{\mathrm{~L}_{1}}+\frac{4}{\mathrm{~L}_{2}}+\frac{4}{\mathrm{~L}_{3}}+\frac{4}{\mathrm{~L}_{4}}$ Here, $\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}_{3}=\mathrm{L}_{4}=\mathrm{L}$ So, $\quad \frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{16}{\mathrm{~L}}$ Hence, $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}}{16}$
AP EAMCET (18.09.2020) Shift-II
Electro Magnetic Induction
154750
The magnetic flux linked with a coil (In Wb) is given by the equation $\phi=5 t^{2}+3 t+16$ The magnitude of induced emf in the coil at the fourth second will be
1 $33 \mathrm{~V}$
2 $43 \mathrm{~V}$
3 $108 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
B Magnetic flux, $\phi=5 \mathrm{t}^{2}+3 \mathrm{t}+16$ Induced emf, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $|\varepsilon|=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+3 \mathrm{t}+16\right)$ $|\varepsilon|=10 \mathrm{t}+3$ At time $\mathrm{t}=4 \mathrm{sec}$ $|\varepsilon|_{\mathrm{t}=4}=10 \times 4+3$ $|\varepsilon|_{\mathrm{t}=4}=43 \text { volt }$
154746
A long straight wire carrying current $I$ and a rectangular frame with side lengths a and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is
1 $\frac{\mu_{0}}{2 \pi} \mathrm{ab}$
2 $\frac{\mu}{4 \pi} \frac{b}{a}$
3 $\frac{\mu_{0} \mathrm{a}}{2 \pi} \ln 2$
4 $\frac{\mu_{0} b}{2 \pi} \ln 2$
Explanation:
D Magnetic field at a distance $r$ from wire $B=\frac{\mu_{0} I}{2 \pi r}$ Flux through the frame is obtained by $\phi_{12}=\int_{\mathrm{a}}^{(\mathrm{a}+\mathrm{a})} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{bdr}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{2 \mathrm{a}}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln \frac{2 \mathrm{a}}{\mathrm{a}}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2$ Mutual induction $\mathrm{M}_{12}=\frac{\phi_{12}}{\mathrm{I}}=\frac{\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2}{\mathrm{I}}$ $\mathrm{M}_{12}=\frac{\mu_{0} \mathrm{~b}}{2 \pi} \ln 2$
TS EAMCET 06.08.2021
Electro Magnetic Induction
154747
A pair of adjacent coils has a mutual inductance of $1.5 \mathrm{H}$. If the current in one coil changes from 0 to $20 \mathrm{~A}$ in 0.5 sec. what is the change of flux linkage with the other coil?
154749
A coil of inductance $L$ is divided into four equal parts and all the parts are connected in parallel. The effective inductance of the combination is
1 $\frac{L}{4}$
2 $\frac{L}{8}$
3 $\frac{L}{16}$
4 $4 L$
Explanation:
C Given, inductance $=\mathrm{L}$ Inductance of each part $=\frac{\mathrm{L}}{4}$ When inductance connected in parallel- $\frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{4}{\mathrm{~L}_{1}}+\frac{4}{\mathrm{~L}_{2}}+\frac{4}{\mathrm{~L}_{3}}+\frac{4}{\mathrm{~L}_{4}}$ Here, $\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}_{3}=\mathrm{L}_{4}=\mathrm{L}$ So, $\quad \frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{16}{\mathrm{~L}}$ Hence, $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}}{16}$
AP EAMCET (18.09.2020) Shift-II
Electro Magnetic Induction
154750
The magnetic flux linked with a coil (In Wb) is given by the equation $\phi=5 t^{2}+3 t+16$ The magnitude of induced emf in the coil at the fourth second will be
1 $33 \mathrm{~V}$
2 $43 \mathrm{~V}$
3 $108 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
B Magnetic flux, $\phi=5 \mathrm{t}^{2}+3 \mathrm{t}+16$ Induced emf, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $|\varepsilon|=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+3 \mathrm{t}+16\right)$ $|\varepsilon|=10 \mathrm{t}+3$ At time $\mathrm{t}=4 \mathrm{sec}$ $|\varepsilon|_{\mathrm{t}=4}=10 \times 4+3$ $|\varepsilon|_{\mathrm{t}=4}=43 \text { volt }$
154746
A long straight wire carrying current $I$ and a rectangular frame with side lengths a and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is
1 $\frac{\mu_{0}}{2 \pi} \mathrm{ab}$
2 $\frac{\mu}{4 \pi} \frac{b}{a}$
3 $\frac{\mu_{0} \mathrm{a}}{2 \pi} \ln 2$
4 $\frac{\mu_{0} b}{2 \pi} \ln 2$
Explanation:
D Magnetic field at a distance $r$ from wire $B=\frac{\mu_{0} I}{2 \pi r}$ Flux through the frame is obtained by $\phi_{12}=\int_{\mathrm{a}}^{(\mathrm{a}+\mathrm{a})} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{bdr}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{2 \mathrm{a}}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln \frac{2 \mathrm{a}}{\mathrm{a}}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2$ Mutual induction $\mathrm{M}_{12}=\frac{\phi_{12}}{\mathrm{I}}=\frac{\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2}{\mathrm{I}}$ $\mathrm{M}_{12}=\frac{\mu_{0} \mathrm{~b}}{2 \pi} \ln 2$
TS EAMCET 06.08.2021
Electro Magnetic Induction
154747
A pair of adjacent coils has a mutual inductance of $1.5 \mathrm{H}$. If the current in one coil changes from 0 to $20 \mathrm{~A}$ in 0.5 sec. what is the change of flux linkage with the other coil?
154749
A coil of inductance $L$ is divided into four equal parts and all the parts are connected in parallel. The effective inductance of the combination is
1 $\frac{L}{4}$
2 $\frac{L}{8}$
3 $\frac{L}{16}$
4 $4 L$
Explanation:
C Given, inductance $=\mathrm{L}$ Inductance of each part $=\frac{\mathrm{L}}{4}$ When inductance connected in parallel- $\frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{4}{\mathrm{~L}_{1}}+\frac{4}{\mathrm{~L}_{2}}+\frac{4}{\mathrm{~L}_{3}}+\frac{4}{\mathrm{~L}_{4}}$ Here, $\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}_{3}=\mathrm{L}_{4}=\mathrm{L}$ So, $\quad \frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{16}{\mathrm{~L}}$ Hence, $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}}{16}$
AP EAMCET (18.09.2020) Shift-II
Electro Magnetic Induction
154750
The magnetic flux linked with a coil (In Wb) is given by the equation $\phi=5 t^{2}+3 t+16$ The magnitude of induced emf in the coil at the fourth second will be
1 $33 \mathrm{~V}$
2 $43 \mathrm{~V}$
3 $108 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
B Magnetic flux, $\phi=5 \mathrm{t}^{2}+3 \mathrm{t}+16$ Induced emf, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $|\varepsilon|=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+3 \mathrm{t}+16\right)$ $|\varepsilon|=10 \mathrm{t}+3$ At time $\mathrm{t}=4 \mathrm{sec}$ $|\varepsilon|_{\mathrm{t}=4}=10 \times 4+3$ $|\varepsilon|_{\mathrm{t}=4}=43 \text { volt }$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154746
A long straight wire carrying current $I$ and a rectangular frame with side lengths a and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is
1 $\frac{\mu_{0}}{2 \pi} \mathrm{ab}$
2 $\frac{\mu}{4 \pi} \frac{b}{a}$
3 $\frac{\mu_{0} \mathrm{a}}{2 \pi} \ln 2$
4 $\frac{\mu_{0} b}{2 \pi} \ln 2$
Explanation:
D Magnetic field at a distance $r$ from wire $B=\frac{\mu_{0} I}{2 \pi r}$ Flux through the frame is obtained by $\phi_{12}=\int_{\mathrm{a}}^{(\mathrm{a}+\mathrm{a})} \frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{bdr}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{2 \mathrm{a}}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln \frac{2 \mathrm{a}}{\mathrm{a}}$ $\phi_{12}=\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2$ Mutual induction $\mathrm{M}_{12}=\frac{\phi_{12}}{\mathrm{I}}=\frac{\frac{\mu_{0} \mathrm{Ib}}{2 \pi} \ln 2}{\mathrm{I}}$ $\mathrm{M}_{12}=\frac{\mu_{0} \mathrm{~b}}{2 \pi} \ln 2$
TS EAMCET 06.08.2021
Electro Magnetic Induction
154747
A pair of adjacent coils has a mutual inductance of $1.5 \mathrm{H}$. If the current in one coil changes from 0 to $20 \mathrm{~A}$ in 0.5 sec. what is the change of flux linkage with the other coil?
154749
A coil of inductance $L$ is divided into four equal parts and all the parts are connected in parallel. The effective inductance of the combination is
1 $\frac{L}{4}$
2 $\frac{L}{8}$
3 $\frac{L}{16}$
4 $4 L$
Explanation:
C Given, inductance $=\mathrm{L}$ Inductance of each part $=\frac{\mathrm{L}}{4}$ When inductance connected in parallel- $\frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{4}{\mathrm{~L}_{1}}+\frac{4}{\mathrm{~L}_{2}}+\frac{4}{\mathrm{~L}_{3}}+\frac{4}{\mathrm{~L}_{4}}$ Here, $\mathrm{L}_{1}=\mathrm{L}_{2}=\mathrm{L}_{3}=\mathrm{L}_{4}=\mathrm{L}$ So, $\quad \frac{1}{\mathrm{~L}_{\mathrm{eq}}}=\frac{16}{\mathrm{~L}}$ Hence, $\mathrm{L}_{\text {eq }}=\frac{\mathrm{L}}{16}$
AP EAMCET (18.09.2020) Shift-II
Electro Magnetic Induction
154750
The magnetic flux linked with a coil (In Wb) is given by the equation $\phi=5 t^{2}+3 t+16$ The magnitude of induced emf in the coil at the fourth second will be
1 $33 \mathrm{~V}$
2 $43 \mathrm{~V}$
3 $108 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
B Magnetic flux, $\phi=5 \mathrm{t}^{2}+3 \mathrm{t}+16$ Induced emf, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $|\varepsilon|=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+3 \mathrm{t}+16\right)$ $|\varepsilon|=10 \mathrm{t}+3$ At time $\mathrm{t}=4 \mathrm{sec}$ $|\varepsilon|_{\mathrm{t}=4}=10 \times 4+3$ $|\varepsilon|_{\mathrm{t}=4}=43 \text { volt }$
