154865
In an AC generator, a coil with $\mathbf{N}$ turns, all of the same area $A$ and total resistance $R$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is
1 NABR $\omega$
2 NAB
3 NABR
4 $\mathrm{NAB} \omega$
Explanation:
D $\because \varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}} =-\frac{\mathrm{d}(\mathrm{NB} \cdot \overrightarrow{\mathrm{A}})}{\mathrm{dt}}$ $\varepsilon =\mathrm{N} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \omega \mathrm{t})$ $\varepsilon =\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum value of emf when $\sin \omega t=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{NBA} \omega$ Where, $\mathrm{N}=$ number of turns $A=\text { area of loop. }$ $B=\text { magnetic field }$ $\omega=\text { frequency of loop. }$
AIIMS-2015
Electro Magnetic Induction
154866
The magnetic flux through a circuit carrying a current of $2.0 \mathrm{~A}$ is 0.8 weber. If the current reduces to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf be:
1 $2.0 \mathrm{~V}$
2 $4.0 \mathrm{~V}$
3 $8.0 \mathrm{~V}$
4 None of the above
Explanation:
A Given that, Current $\mathrm{I}_{1}=2 \mathrm{~A}$ Reduced current $\mathrm{I}_{2}=1.5 \mathrm{~A}$ Magnetic flux $\phi_{1}=0.8 \mathrm{~Wb}$ Time, $\Delta \mathrm{t}=0.1 \mathrm{sec}$ $\because \quad \phi_{1}=\mathrm{LI}_{1}$ $\mathrm{L}=\frac{\phi_{1}}{\mathrm{I}_{1}}=\frac{0.8}{2}=0.4$ $\phi_{2}=\mathrm{L} \mathrm{I}_{2}$ $\phi_{2}=0.4 \times 1.5=0.6 \mathrm{~Wb}$ $\operatorname{emf} \varepsilon =\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\therefore \quad \operatorname{emf} \varepsilon =\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{0.8-0.6}{0.1}$ $\operatorname{emf} \varepsilon =2 \mathrm{~V}$
AIIMS-2011
Electro Magnetic Induction
154867
A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is
1 a
2 b
3 c
4 d
Explanation:
A We know that $\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ In the beginning $\mathrm{d} \phi$ will be positive (at time $\frac{\mathrm{T}}{4}$ ), Therefore it will be negative during the time period from $\frac{\mathrm{T}}{4}$ to $\frac{\mathrm{T}}{2}$. And, during the period $\frac{T}{2}$ to $\frac{3 T}{4}$, it will be positive again and in the last $\frac{\mathrm{T}}{4}$ time it will be negative. Hence, option (a) illustrates the correct time variation of the magnitude of emf generated across the coil during one cycle.
AIIMS-2005
Electro Magnetic Induction
154868
In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f induced in the coil is :
1 $5 \mathrm{~V}$
2 $-5 \mathrm{~V}$
3 $-10 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, Self inductance $\mathrm{L}=5$ Henry, Rate of change of current $\frac{\mathrm{dI}}{\mathrm{dt}}=2 \mathrm{~A} / \mathrm{sec}$ $\because \quad \quad \operatorname{emf} \varepsilon=-\mathrm{L} \times \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=-5 \times 2$ emf $\varepsilon=-10 \mathrm{~V}$
AIIMS-1997
Electro Magnetic Induction
154869
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is-
1 halved
2 doubled
3 quadrupled
4 reduced to one-fourth
Explanation:
B As we know that, $\mathrm{M}=\mathrm{K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ If $\mathrm{L}_{1}$ becomes $2 \mathrm{~L}_{1}$ and $\mathrm{L}_{2}$ becomes $2 \mathrm{~L}_{2}$ Then, $\mathrm{M}^{\prime} =\mathrm{K} \sqrt{2 \mathrm{~L}_{1} 2 \mathrm{~L}_{2}}$ $=\mathrm{K} \sqrt{4 \mathrm{~L}_{1} \mathrm{~L}_{2}}=2 \mathrm{~K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\mathrm{M}^{\prime} =2 \mathrm{M}$ Where ' $\mathrm{K}$ ' is constant and $\mathrm{M}$ is mutual inductance
154865
In an AC generator, a coil with $\mathbf{N}$ turns, all of the same area $A$ and total resistance $R$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is
1 NABR $\omega$
2 NAB
3 NABR
4 $\mathrm{NAB} \omega$
Explanation:
D $\because \varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}} =-\frac{\mathrm{d}(\mathrm{NB} \cdot \overrightarrow{\mathrm{A}})}{\mathrm{dt}}$ $\varepsilon =\mathrm{N} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \omega \mathrm{t})$ $\varepsilon =\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum value of emf when $\sin \omega t=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{NBA} \omega$ Where, $\mathrm{N}=$ number of turns $A=\text { area of loop. }$ $B=\text { magnetic field }$ $\omega=\text { frequency of loop. }$
AIIMS-2015
Electro Magnetic Induction
154866
The magnetic flux through a circuit carrying a current of $2.0 \mathrm{~A}$ is 0.8 weber. If the current reduces to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf be:
1 $2.0 \mathrm{~V}$
2 $4.0 \mathrm{~V}$
3 $8.0 \mathrm{~V}$
4 None of the above
Explanation:
A Given that, Current $\mathrm{I}_{1}=2 \mathrm{~A}$ Reduced current $\mathrm{I}_{2}=1.5 \mathrm{~A}$ Magnetic flux $\phi_{1}=0.8 \mathrm{~Wb}$ Time, $\Delta \mathrm{t}=0.1 \mathrm{sec}$ $\because \quad \phi_{1}=\mathrm{LI}_{1}$ $\mathrm{L}=\frac{\phi_{1}}{\mathrm{I}_{1}}=\frac{0.8}{2}=0.4$ $\phi_{2}=\mathrm{L} \mathrm{I}_{2}$ $\phi_{2}=0.4 \times 1.5=0.6 \mathrm{~Wb}$ $\operatorname{emf} \varepsilon =\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\therefore \quad \operatorname{emf} \varepsilon =\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{0.8-0.6}{0.1}$ $\operatorname{emf} \varepsilon =2 \mathrm{~V}$
AIIMS-2011
Electro Magnetic Induction
154867
A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is
1 a
2 b
3 c
4 d
Explanation:
A We know that $\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ In the beginning $\mathrm{d} \phi$ will be positive (at time $\frac{\mathrm{T}}{4}$ ), Therefore it will be negative during the time period from $\frac{\mathrm{T}}{4}$ to $\frac{\mathrm{T}}{2}$. And, during the period $\frac{T}{2}$ to $\frac{3 T}{4}$, it will be positive again and in the last $\frac{\mathrm{T}}{4}$ time it will be negative. Hence, option (a) illustrates the correct time variation of the magnitude of emf generated across the coil during one cycle.
AIIMS-2005
Electro Magnetic Induction
154868
In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f induced in the coil is :
1 $5 \mathrm{~V}$
2 $-5 \mathrm{~V}$
3 $-10 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, Self inductance $\mathrm{L}=5$ Henry, Rate of change of current $\frac{\mathrm{dI}}{\mathrm{dt}}=2 \mathrm{~A} / \mathrm{sec}$ $\because \quad \quad \operatorname{emf} \varepsilon=-\mathrm{L} \times \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=-5 \times 2$ emf $\varepsilon=-10 \mathrm{~V}$
AIIMS-1997
Electro Magnetic Induction
154869
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is-
1 halved
2 doubled
3 quadrupled
4 reduced to one-fourth
Explanation:
B As we know that, $\mathrm{M}=\mathrm{K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ If $\mathrm{L}_{1}$ becomes $2 \mathrm{~L}_{1}$ and $\mathrm{L}_{2}$ becomes $2 \mathrm{~L}_{2}$ Then, $\mathrm{M}^{\prime} =\mathrm{K} \sqrt{2 \mathrm{~L}_{1} 2 \mathrm{~L}_{2}}$ $=\mathrm{K} \sqrt{4 \mathrm{~L}_{1} \mathrm{~L}_{2}}=2 \mathrm{~K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\mathrm{M}^{\prime} =2 \mathrm{M}$ Where ' $\mathrm{K}$ ' is constant and $\mathrm{M}$ is mutual inductance
154865
In an AC generator, a coil with $\mathbf{N}$ turns, all of the same area $A$ and total resistance $R$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is
1 NABR $\omega$
2 NAB
3 NABR
4 $\mathrm{NAB} \omega$
Explanation:
D $\because \varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}} =-\frac{\mathrm{d}(\mathrm{NB} \cdot \overrightarrow{\mathrm{A}})}{\mathrm{dt}}$ $\varepsilon =\mathrm{N} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \omega \mathrm{t})$ $\varepsilon =\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum value of emf when $\sin \omega t=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{NBA} \omega$ Where, $\mathrm{N}=$ number of turns $A=\text { area of loop. }$ $B=\text { magnetic field }$ $\omega=\text { frequency of loop. }$
AIIMS-2015
Electro Magnetic Induction
154866
The magnetic flux through a circuit carrying a current of $2.0 \mathrm{~A}$ is 0.8 weber. If the current reduces to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf be:
1 $2.0 \mathrm{~V}$
2 $4.0 \mathrm{~V}$
3 $8.0 \mathrm{~V}$
4 None of the above
Explanation:
A Given that, Current $\mathrm{I}_{1}=2 \mathrm{~A}$ Reduced current $\mathrm{I}_{2}=1.5 \mathrm{~A}$ Magnetic flux $\phi_{1}=0.8 \mathrm{~Wb}$ Time, $\Delta \mathrm{t}=0.1 \mathrm{sec}$ $\because \quad \phi_{1}=\mathrm{LI}_{1}$ $\mathrm{L}=\frac{\phi_{1}}{\mathrm{I}_{1}}=\frac{0.8}{2}=0.4$ $\phi_{2}=\mathrm{L} \mathrm{I}_{2}$ $\phi_{2}=0.4 \times 1.5=0.6 \mathrm{~Wb}$ $\operatorname{emf} \varepsilon =\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\therefore \quad \operatorname{emf} \varepsilon =\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{0.8-0.6}{0.1}$ $\operatorname{emf} \varepsilon =2 \mathrm{~V}$
AIIMS-2011
Electro Magnetic Induction
154867
A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is
1 a
2 b
3 c
4 d
Explanation:
A We know that $\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ In the beginning $\mathrm{d} \phi$ will be positive (at time $\frac{\mathrm{T}}{4}$ ), Therefore it will be negative during the time period from $\frac{\mathrm{T}}{4}$ to $\frac{\mathrm{T}}{2}$. And, during the period $\frac{T}{2}$ to $\frac{3 T}{4}$, it will be positive again and in the last $\frac{\mathrm{T}}{4}$ time it will be negative. Hence, option (a) illustrates the correct time variation of the magnitude of emf generated across the coil during one cycle.
AIIMS-2005
Electro Magnetic Induction
154868
In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f induced in the coil is :
1 $5 \mathrm{~V}$
2 $-5 \mathrm{~V}$
3 $-10 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, Self inductance $\mathrm{L}=5$ Henry, Rate of change of current $\frac{\mathrm{dI}}{\mathrm{dt}}=2 \mathrm{~A} / \mathrm{sec}$ $\because \quad \quad \operatorname{emf} \varepsilon=-\mathrm{L} \times \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=-5 \times 2$ emf $\varepsilon=-10 \mathrm{~V}$
AIIMS-1997
Electro Magnetic Induction
154869
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is-
1 halved
2 doubled
3 quadrupled
4 reduced to one-fourth
Explanation:
B As we know that, $\mathrm{M}=\mathrm{K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ If $\mathrm{L}_{1}$ becomes $2 \mathrm{~L}_{1}$ and $\mathrm{L}_{2}$ becomes $2 \mathrm{~L}_{2}$ Then, $\mathrm{M}^{\prime} =\mathrm{K} \sqrt{2 \mathrm{~L}_{1} 2 \mathrm{~L}_{2}}$ $=\mathrm{K} \sqrt{4 \mathrm{~L}_{1} \mathrm{~L}_{2}}=2 \mathrm{~K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\mathrm{M}^{\prime} =2 \mathrm{M}$ Where ' $\mathrm{K}$ ' is constant and $\mathrm{M}$ is mutual inductance
154865
In an AC generator, a coil with $\mathbf{N}$ turns, all of the same area $A$ and total resistance $R$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is
1 NABR $\omega$
2 NAB
3 NABR
4 $\mathrm{NAB} \omega$
Explanation:
D $\because \varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}} =-\frac{\mathrm{d}(\mathrm{NB} \cdot \overrightarrow{\mathrm{A}})}{\mathrm{dt}}$ $\varepsilon =\mathrm{N} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \omega \mathrm{t})$ $\varepsilon =\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum value of emf when $\sin \omega t=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{NBA} \omega$ Where, $\mathrm{N}=$ number of turns $A=\text { area of loop. }$ $B=\text { magnetic field }$ $\omega=\text { frequency of loop. }$
AIIMS-2015
Electro Magnetic Induction
154866
The magnetic flux through a circuit carrying a current of $2.0 \mathrm{~A}$ is 0.8 weber. If the current reduces to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf be:
1 $2.0 \mathrm{~V}$
2 $4.0 \mathrm{~V}$
3 $8.0 \mathrm{~V}$
4 None of the above
Explanation:
A Given that, Current $\mathrm{I}_{1}=2 \mathrm{~A}$ Reduced current $\mathrm{I}_{2}=1.5 \mathrm{~A}$ Magnetic flux $\phi_{1}=0.8 \mathrm{~Wb}$ Time, $\Delta \mathrm{t}=0.1 \mathrm{sec}$ $\because \quad \phi_{1}=\mathrm{LI}_{1}$ $\mathrm{L}=\frac{\phi_{1}}{\mathrm{I}_{1}}=\frac{0.8}{2}=0.4$ $\phi_{2}=\mathrm{L} \mathrm{I}_{2}$ $\phi_{2}=0.4 \times 1.5=0.6 \mathrm{~Wb}$ $\operatorname{emf} \varepsilon =\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\therefore \quad \operatorname{emf} \varepsilon =\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{0.8-0.6}{0.1}$ $\operatorname{emf} \varepsilon =2 \mathrm{~V}$
AIIMS-2011
Electro Magnetic Induction
154867
A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is
1 a
2 b
3 c
4 d
Explanation:
A We know that $\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ In the beginning $\mathrm{d} \phi$ will be positive (at time $\frac{\mathrm{T}}{4}$ ), Therefore it will be negative during the time period from $\frac{\mathrm{T}}{4}$ to $\frac{\mathrm{T}}{2}$. And, during the period $\frac{T}{2}$ to $\frac{3 T}{4}$, it will be positive again and in the last $\frac{\mathrm{T}}{4}$ time it will be negative. Hence, option (a) illustrates the correct time variation of the magnitude of emf generated across the coil during one cycle.
AIIMS-2005
Electro Magnetic Induction
154868
In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f induced in the coil is :
1 $5 \mathrm{~V}$
2 $-5 \mathrm{~V}$
3 $-10 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, Self inductance $\mathrm{L}=5$ Henry, Rate of change of current $\frac{\mathrm{dI}}{\mathrm{dt}}=2 \mathrm{~A} / \mathrm{sec}$ $\because \quad \quad \operatorname{emf} \varepsilon=-\mathrm{L} \times \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=-5 \times 2$ emf $\varepsilon=-10 \mathrm{~V}$
AIIMS-1997
Electro Magnetic Induction
154869
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is-
1 halved
2 doubled
3 quadrupled
4 reduced to one-fourth
Explanation:
B As we know that, $\mathrm{M}=\mathrm{K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ If $\mathrm{L}_{1}$ becomes $2 \mathrm{~L}_{1}$ and $\mathrm{L}_{2}$ becomes $2 \mathrm{~L}_{2}$ Then, $\mathrm{M}^{\prime} =\mathrm{K} \sqrt{2 \mathrm{~L}_{1} 2 \mathrm{~L}_{2}}$ $=\mathrm{K} \sqrt{4 \mathrm{~L}_{1} \mathrm{~L}_{2}}=2 \mathrm{~K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\mathrm{M}^{\prime} =2 \mathrm{M}$ Where ' $\mathrm{K}$ ' is constant and $\mathrm{M}$ is mutual inductance
154865
In an AC generator, a coil with $\mathbf{N}$ turns, all of the same area $A$ and total resistance $R$, rotates with frequency $\omega$ in a magnetic field $B$. The maximum value of emf generated in the coil is
1 NABR $\omega$
2 NAB
3 NABR
4 $\mathrm{NAB} \omega$
Explanation:
D $\because \varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}} =-\frac{\mathrm{d}(\mathrm{NB} \cdot \overrightarrow{\mathrm{A}})}{\mathrm{dt}}$ $\varepsilon =\mathrm{N} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA} \cos \omega \mathrm{t})$ $\varepsilon =\mathrm{NBA} \omega \sin \omega \mathrm{t}$ For maximum value of emf when $\sin \omega t=1$ $\therefore \quad \varepsilon_{\max }=\mathrm{NBA} \omega$ Where, $\mathrm{N}=$ number of turns $A=\text { area of loop. }$ $B=\text { magnetic field }$ $\omega=\text { frequency of loop. }$
AIIMS-2015
Electro Magnetic Induction
154866
The magnetic flux through a circuit carrying a current of $2.0 \mathrm{~A}$ is 0.8 weber. If the current reduces to $1.5 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf be:
1 $2.0 \mathrm{~V}$
2 $4.0 \mathrm{~V}$
3 $8.0 \mathrm{~V}$
4 None of the above
Explanation:
A Given that, Current $\mathrm{I}_{1}=2 \mathrm{~A}$ Reduced current $\mathrm{I}_{2}=1.5 \mathrm{~A}$ Magnetic flux $\phi_{1}=0.8 \mathrm{~Wb}$ Time, $\Delta \mathrm{t}=0.1 \mathrm{sec}$ $\because \quad \phi_{1}=\mathrm{LI}_{1}$ $\mathrm{L}=\frac{\phi_{1}}{\mathrm{I}_{1}}=\frac{0.8}{2}=0.4$ $\phi_{2}=\mathrm{L} \mathrm{I}_{2}$ $\phi_{2}=0.4 \times 1.5=0.6 \mathrm{~Wb}$ $\operatorname{emf} \varepsilon =\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\therefore \quad \operatorname{emf} \varepsilon =\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{0.8-0.6}{0.1}$ $\operatorname{emf} \varepsilon =2 \mathrm{~V}$
AIIMS-2011
Electro Magnetic Induction
154867
A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is
1 a
2 b
3 c
4 d
Explanation:
A We know that $\varepsilon=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ In the beginning $\mathrm{d} \phi$ will be positive (at time $\frac{\mathrm{T}}{4}$ ), Therefore it will be negative during the time period from $\frac{\mathrm{T}}{4}$ to $\frac{\mathrm{T}}{2}$. And, during the period $\frac{T}{2}$ to $\frac{3 T}{4}$, it will be positive again and in the last $\frac{\mathrm{T}}{4}$ time it will be negative. Hence, option (a) illustrates the correct time variation of the magnitude of emf generated across the coil during one cycle.
AIIMS-2005
Electro Magnetic Induction
154868
In a coil of self inductance of 5 henry, the rate of change of current is 2 ampere per second, the e.m.f induced in the coil is :
1 $5 \mathrm{~V}$
2 $-5 \mathrm{~V}$
3 $-10 \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
C Given that, Self inductance $\mathrm{L}=5$ Henry, Rate of change of current $\frac{\mathrm{dI}}{\mathrm{dt}}=2 \mathrm{~A} / \mathrm{sec}$ $\because \quad \quad \operatorname{emf} \varepsilon=-\mathrm{L} \times \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=-5 \times 2$ emf $\varepsilon=-10 \mathrm{~V}$
AIIMS-1997
Electro Magnetic Induction
154869
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is-
1 halved
2 doubled
3 quadrupled
4 reduced to one-fourth
Explanation:
B As we know that, $\mathrm{M}=\mathrm{K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ If $\mathrm{L}_{1}$ becomes $2 \mathrm{~L}_{1}$ and $\mathrm{L}_{2}$ becomes $2 \mathrm{~L}_{2}$ Then, $\mathrm{M}^{\prime} =\mathrm{K} \sqrt{2 \mathrm{~L}_{1} 2 \mathrm{~L}_{2}}$ $=\mathrm{K} \sqrt{4 \mathrm{~L}_{1} \mathrm{~L}_{2}}=2 \mathrm{~K} \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\mathrm{M}^{\prime} =2 \mathrm{M}$ Where ' $\mathrm{K}$ ' is constant and $\mathrm{M}$ is mutual inductance
