154762
The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 \mathrm{~A}$ in $0.006 \mathrm{~s}$, the e.m.f. induced in the secondary coil at that instant is
1 $120 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $180 \mathrm{~V}$
4 $300 \mathrm{~V}$
Explanation:
D Given, Mutual inductance $\mathrm{M}=0.09 \mathrm{H}$ Change in current $|\Delta I|=I_{1}-I_{2}=0-20$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.006 \mathrm{~s}$ $\because \quad \varepsilon=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}=\frac{0.09 \times 20}{0.006}$ $\varepsilon=300 \mathrm{~V}$
MHT-CET 2020
Electro Magnetic Induction
154763
Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $\left(R_{2}>R_{1}\right)$. If a current is passed through coil $B$, the coefficient of mutual inductance between the coils is proportional to
1 $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
2 $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
3 $\frac{1}{R_{1} R_{2}}$
4 $\mathrm{R}_{1} \mathrm{R}_{2}$
Explanation:
A : For bigger loop- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}}$ So, the flux associated within inner loop $\phi_{2}=\text { B. } \mathrm{A}_{1}$ $\phi_{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \pi \mathrm{R}_{1}^{2}$ $\because \quad \phi_{2}=\mathrm{MI}_{1}$ $\therefore \quad \mathrm{M}=\frac{\phi_{2}}{\mathrm{I}_{1}}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \frac{\pi \mathrm{R}_{1}^{2}}{\mathrm{I}_{1}}$ $\mathrm{M}=\frac{\mu_{\mathrm{o}} \pi \mathrm{R}_{1}^{2}}{2 \mathrm{R}_{2}}$ $\therefore \quad \mathrm{M} \propto \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
MHT-CET 2020
Electro Magnetic Induction
154764
A current $I=10 \sin (50 \pi t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 \pi$ volt in the second coil. The mutual inductance between the coils is
154765
A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)
D Given, Average radius of toroidal solenoid $=\mathrm{R}$. Number of turns $=\mathrm{N}$ Area of crosesoction $=\mathrm{A}$ Then, self inductance of solenoide, $=\frac{\mu_{\mathrm{o}} \mathrm{N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
MHT-CET 2020
Electro Magnetic Induction
154767
Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$. If the current in the primary is $I=I_{0} \sin \omega t$, then the maximum value of e.m.f. induced in coil $Q$ is
1 $\frac{M}{I_{0} \omega}$
2 $\mathrm{I}_{0} \mathrm{M} \omega$
3 $\frac{\mathrm{I}_{0}}{\mathrm{M} \omega}$
4 $\frac{\omega}{\mathrm{I}_{0} \mathrm{M}}$
Explanation:
B Given $\mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \omega \mathrm{t}$ Mutual inductance $=\mathrm{M}$ Induced emf in the coil $\varepsilon =M \frac{d I}{d t}$ $=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)$ $\varepsilon =M \omega I_{0} \cos \omega t$ For maximum value of $\varepsilon, \cos \omega \mathrm{t}=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{M} \mathrm{I}_{0} \omega$
154762
The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 \mathrm{~A}$ in $0.006 \mathrm{~s}$, the e.m.f. induced in the secondary coil at that instant is
1 $120 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $180 \mathrm{~V}$
4 $300 \mathrm{~V}$
Explanation:
D Given, Mutual inductance $\mathrm{M}=0.09 \mathrm{H}$ Change in current $|\Delta I|=I_{1}-I_{2}=0-20$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.006 \mathrm{~s}$ $\because \quad \varepsilon=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}=\frac{0.09 \times 20}{0.006}$ $\varepsilon=300 \mathrm{~V}$
MHT-CET 2020
Electro Magnetic Induction
154763
Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $\left(R_{2}>R_{1}\right)$. If a current is passed through coil $B$, the coefficient of mutual inductance between the coils is proportional to
1 $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
2 $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
3 $\frac{1}{R_{1} R_{2}}$
4 $\mathrm{R}_{1} \mathrm{R}_{2}$
Explanation:
A : For bigger loop- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}}$ So, the flux associated within inner loop $\phi_{2}=\text { B. } \mathrm{A}_{1}$ $\phi_{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \pi \mathrm{R}_{1}^{2}$ $\because \quad \phi_{2}=\mathrm{MI}_{1}$ $\therefore \quad \mathrm{M}=\frac{\phi_{2}}{\mathrm{I}_{1}}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \frac{\pi \mathrm{R}_{1}^{2}}{\mathrm{I}_{1}}$ $\mathrm{M}=\frac{\mu_{\mathrm{o}} \pi \mathrm{R}_{1}^{2}}{2 \mathrm{R}_{2}}$ $\therefore \quad \mathrm{M} \propto \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
MHT-CET 2020
Electro Magnetic Induction
154764
A current $I=10 \sin (50 \pi t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 \pi$ volt in the second coil. The mutual inductance between the coils is
154765
A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)
D Given, Average radius of toroidal solenoid $=\mathrm{R}$. Number of turns $=\mathrm{N}$ Area of crosesoction $=\mathrm{A}$ Then, self inductance of solenoide, $=\frac{\mu_{\mathrm{o}} \mathrm{N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
MHT-CET 2020
Electro Magnetic Induction
154767
Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$. If the current in the primary is $I=I_{0} \sin \omega t$, then the maximum value of e.m.f. induced in coil $Q$ is
1 $\frac{M}{I_{0} \omega}$
2 $\mathrm{I}_{0} \mathrm{M} \omega$
3 $\frac{\mathrm{I}_{0}}{\mathrm{M} \omega}$
4 $\frac{\omega}{\mathrm{I}_{0} \mathrm{M}}$
Explanation:
B Given $\mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \omega \mathrm{t}$ Mutual inductance $=\mathrm{M}$ Induced emf in the coil $\varepsilon =M \frac{d I}{d t}$ $=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)$ $\varepsilon =M \omega I_{0} \cos \omega t$ For maximum value of $\varepsilon, \cos \omega \mathrm{t}=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{M} \mathrm{I}_{0} \omega$
154762
The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 \mathrm{~A}$ in $0.006 \mathrm{~s}$, the e.m.f. induced in the secondary coil at that instant is
1 $120 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $180 \mathrm{~V}$
4 $300 \mathrm{~V}$
Explanation:
D Given, Mutual inductance $\mathrm{M}=0.09 \mathrm{H}$ Change in current $|\Delta I|=I_{1}-I_{2}=0-20$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.006 \mathrm{~s}$ $\because \quad \varepsilon=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}=\frac{0.09 \times 20}{0.006}$ $\varepsilon=300 \mathrm{~V}$
MHT-CET 2020
Electro Magnetic Induction
154763
Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $\left(R_{2}>R_{1}\right)$. If a current is passed through coil $B$, the coefficient of mutual inductance between the coils is proportional to
1 $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
2 $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
3 $\frac{1}{R_{1} R_{2}}$
4 $\mathrm{R}_{1} \mathrm{R}_{2}$
Explanation:
A : For bigger loop- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}}$ So, the flux associated within inner loop $\phi_{2}=\text { B. } \mathrm{A}_{1}$ $\phi_{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \pi \mathrm{R}_{1}^{2}$ $\because \quad \phi_{2}=\mathrm{MI}_{1}$ $\therefore \quad \mathrm{M}=\frac{\phi_{2}}{\mathrm{I}_{1}}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \frac{\pi \mathrm{R}_{1}^{2}}{\mathrm{I}_{1}}$ $\mathrm{M}=\frac{\mu_{\mathrm{o}} \pi \mathrm{R}_{1}^{2}}{2 \mathrm{R}_{2}}$ $\therefore \quad \mathrm{M} \propto \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
MHT-CET 2020
Electro Magnetic Induction
154764
A current $I=10 \sin (50 \pi t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 \pi$ volt in the second coil. The mutual inductance between the coils is
154765
A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)
D Given, Average radius of toroidal solenoid $=\mathrm{R}$. Number of turns $=\mathrm{N}$ Area of crosesoction $=\mathrm{A}$ Then, self inductance of solenoide, $=\frac{\mu_{\mathrm{o}} \mathrm{N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
MHT-CET 2020
Electro Magnetic Induction
154767
Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$. If the current in the primary is $I=I_{0} \sin \omega t$, then the maximum value of e.m.f. induced in coil $Q$ is
1 $\frac{M}{I_{0} \omega}$
2 $\mathrm{I}_{0} \mathrm{M} \omega$
3 $\frac{\mathrm{I}_{0}}{\mathrm{M} \omega}$
4 $\frac{\omega}{\mathrm{I}_{0} \mathrm{M}}$
Explanation:
B Given $\mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \omega \mathrm{t}$ Mutual inductance $=\mathrm{M}$ Induced emf in the coil $\varepsilon =M \frac{d I}{d t}$ $=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)$ $\varepsilon =M \omega I_{0} \cos \omega t$ For maximum value of $\varepsilon, \cos \omega \mathrm{t}=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{M} \mathrm{I}_{0} \omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154762
The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 \mathrm{~A}$ in $0.006 \mathrm{~s}$, the e.m.f. induced in the secondary coil at that instant is
1 $120 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $180 \mathrm{~V}$
4 $300 \mathrm{~V}$
Explanation:
D Given, Mutual inductance $\mathrm{M}=0.09 \mathrm{H}$ Change in current $|\Delta I|=I_{1}-I_{2}=0-20$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.006 \mathrm{~s}$ $\because \quad \varepsilon=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}=\frac{0.09 \times 20}{0.006}$ $\varepsilon=300 \mathrm{~V}$
MHT-CET 2020
Electro Magnetic Induction
154763
Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $\left(R_{2}>R_{1}\right)$. If a current is passed through coil $B$, the coefficient of mutual inductance between the coils is proportional to
1 $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
2 $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
3 $\frac{1}{R_{1} R_{2}}$
4 $\mathrm{R}_{1} \mathrm{R}_{2}$
Explanation:
A : For bigger loop- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}}$ So, the flux associated within inner loop $\phi_{2}=\text { B. } \mathrm{A}_{1}$ $\phi_{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \pi \mathrm{R}_{1}^{2}$ $\because \quad \phi_{2}=\mathrm{MI}_{1}$ $\therefore \quad \mathrm{M}=\frac{\phi_{2}}{\mathrm{I}_{1}}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \frac{\pi \mathrm{R}_{1}^{2}}{\mathrm{I}_{1}}$ $\mathrm{M}=\frac{\mu_{\mathrm{o}} \pi \mathrm{R}_{1}^{2}}{2 \mathrm{R}_{2}}$ $\therefore \quad \mathrm{M} \propto \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
MHT-CET 2020
Electro Magnetic Induction
154764
A current $I=10 \sin (50 \pi t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 \pi$ volt in the second coil. The mutual inductance between the coils is
154765
A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)
D Given, Average radius of toroidal solenoid $=\mathrm{R}$. Number of turns $=\mathrm{N}$ Area of crosesoction $=\mathrm{A}$ Then, self inductance of solenoide, $=\frac{\mu_{\mathrm{o}} \mathrm{N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
MHT-CET 2020
Electro Magnetic Induction
154767
Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$. If the current in the primary is $I=I_{0} \sin \omega t$, then the maximum value of e.m.f. induced in coil $Q$ is
1 $\frac{M}{I_{0} \omega}$
2 $\mathrm{I}_{0} \mathrm{M} \omega$
3 $\frac{\mathrm{I}_{0}}{\mathrm{M} \omega}$
4 $\frac{\omega}{\mathrm{I}_{0} \mathrm{M}}$
Explanation:
B Given $\mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \omega \mathrm{t}$ Mutual inductance $=\mathrm{M}$ Induced emf in the coil $\varepsilon =M \frac{d I}{d t}$ $=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)$ $\varepsilon =M \omega I_{0} \cos \omega t$ For maximum value of $\varepsilon, \cos \omega \mathrm{t}=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{M} \mathrm{I}_{0} \omega$
154762
The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 \mathrm{~A}$ in $0.006 \mathrm{~s}$, the e.m.f. induced in the secondary coil at that instant is
1 $120 \mathrm{~V}$
2 $200 \mathrm{~V}$
3 $180 \mathrm{~V}$
4 $300 \mathrm{~V}$
Explanation:
D Given, Mutual inductance $\mathrm{M}=0.09 \mathrm{H}$ Change in current $|\Delta I|=I_{1}-I_{2}=0-20$ $\Delta \mathrm{I}=20 \mathrm{~A}$ Time $\Delta \mathrm{t}=0.006 \mathrm{~s}$ $\because \quad \varepsilon=\frac{\mathrm{M} \Delta \mathrm{I}}{\Delta \mathrm{t}}=\frac{0.09 \times 20}{0.006}$ $\varepsilon=300 \mathrm{~V}$
MHT-CET 2020
Electro Magnetic Induction
154763
Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $\left(R_{2}>R_{1}\right)$. If a current is passed through coil $B$, the coefficient of mutual inductance between the coils is proportional to
1 $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
2 $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
3 $\frac{1}{R_{1} R_{2}}$
4 $\mathrm{R}_{1} \mathrm{R}_{2}$
Explanation:
A : For bigger loop- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}}$ So, the flux associated within inner loop $\phi_{2}=\text { B. } \mathrm{A}_{1}$ $\phi_{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \pi \mathrm{R}_{1}^{2}$ $\because \quad \phi_{2}=\mathrm{MI}_{1}$ $\therefore \quad \mathrm{M}=\frac{\phi_{2}}{\mathrm{I}_{1}}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{R}_{2}} \times \frac{\pi \mathrm{R}_{1}^{2}}{\mathrm{I}_{1}}$ $\mathrm{M}=\frac{\mu_{\mathrm{o}} \pi \mathrm{R}_{1}^{2}}{2 \mathrm{R}_{2}}$ $\therefore \quad \mathrm{M} \propto \frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
MHT-CET 2020
Electro Magnetic Induction
154764
A current $I=10 \sin (50 \pi t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 \pi$ volt in the second coil. The mutual inductance between the coils is
154765
A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)
D Given, Average radius of toroidal solenoid $=\mathrm{R}$. Number of turns $=\mathrm{N}$ Area of crosesoction $=\mathrm{A}$ Then, self inductance of solenoide, $=\frac{\mu_{\mathrm{o}} \mathrm{N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}$
MHT-CET 2020
Electro Magnetic Induction
154767
Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$. If the current in the primary is $I=I_{0} \sin \omega t$, then the maximum value of e.m.f. induced in coil $Q$ is
1 $\frac{M}{I_{0} \omega}$
2 $\mathrm{I}_{0} \mathrm{M} \omega$
3 $\frac{\mathrm{I}_{0}}{\mathrm{M} \omega}$
4 $\frac{\omega}{\mathrm{I}_{0} \mathrm{M}}$
Explanation:
B Given $\mathrm{I}=\mathrm{I}_{\mathrm{o}} \sin \omega \mathrm{t}$ Mutual inductance $=\mathrm{M}$ Induced emf in the coil $\varepsilon =M \frac{d I}{d t}$ $=M \frac{d}{d t}\left(I_{0} \sin \omega t\right)$ $\varepsilon =M \omega I_{0} \cos \omega t$ For maximum value of $\varepsilon, \cos \omega \mathrm{t}=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{M} \mathrm{I}_{0} \omega$
