154852
The current in a coil changes from $1 \mathrm{~mA}$ to 5 $\mathrm{mA}$ in $4 \mathrm{~ms}$. If the coefficient of self-inductance of the coil is $10 \mathrm{mH}$, the magnitude of "selfinduced" emf is
154853
The current in a coil changes from $4 \mathrm{~A}$ to zero in $1 \mathrm{~ms}$. The average emf induced in the coil is $4 \mathrm{~V}$. The self inductance, in $\mathrm{mH}$, between the coil is
1 2.5
2 0.25
3 25
4 1
Explanation:
D Given that, $\mathrm{I}_1=4 \mathrm{~A}$ $\text { And } \mathrm{I}_2=0 \mathrm{~A}$ $\mathrm{t}=1 \mathrm{~m} \mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ $=4 \mathrm{~V}$ We know, $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \mathrm{L} =\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L} =\frac{4}{(4-0) / 1 \times 10^{-3}}$ $\mathrm{~L} =10^{-3} \mathrm{H}$ $\text { Or } \mathrm{L} =1 \mathrm{mH}$
COMEDK 2011
Electro Magnetic Induction
154854
A current of 2.5 A flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is
1 $0.5 \mathrm{~Wb}$
2 $12.5 \mathrm{~Wb}$
3 zero
4 $2 \mathrm{~Wb}$
Explanation:
B Given- \(\mathrm{I}=2.5 \mathrm{~A} . \quad \mathrm{L}=5 \mathrm{H} \text {. }\) Magnetic flux,\(\quad \phi=\) ? \(\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}\) \(\mathrm{d} \phi=\mathrm{L} \cdot \mathrm{dI}\) or \(\phi=\mathrm{LI}\) \(\phi=5 \times 2.5\) \(\phi=12.5 \mathrm{~Wb}\)
COMEDK 2013
Electro Magnetic Induction
154857
If a change in current of $0.01 \mathrm{~A}$ in one coil produces a change in magnetic flux of $2 \times 10^{-2}$ weber in another coil, then the mutual inductance between coils is
1 0
2 $0.5 \mathrm{H}$
3 $2 \mathrm{H}$
4 $3 \mathrm{H}$
Explanation:
C Given that, Change in current $\Delta \mathrm{I}=0.01 \mathrm{~A}$, Magnetic flux $\Delta \phi=2 \times 10^{-2} \mathrm{~Wb}$ $\mathrm{M}=\text { ? }$ We know, $\Delta \phi=\mathrm{MI}$ $\mathrm{M}=\frac{\Delta \phi}{\Delta \mathrm{I}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \mathrm{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electro Magnetic Induction
154852
The current in a coil changes from $1 \mathrm{~mA}$ to 5 $\mathrm{mA}$ in $4 \mathrm{~ms}$. If the coefficient of self-inductance of the coil is $10 \mathrm{mH}$, the magnitude of "selfinduced" emf is
154853
The current in a coil changes from $4 \mathrm{~A}$ to zero in $1 \mathrm{~ms}$. The average emf induced in the coil is $4 \mathrm{~V}$. The self inductance, in $\mathrm{mH}$, between the coil is
1 2.5
2 0.25
3 25
4 1
Explanation:
D Given that, $\mathrm{I}_1=4 \mathrm{~A}$ $\text { And } \mathrm{I}_2=0 \mathrm{~A}$ $\mathrm{t}=1 \mathrm{~m} \mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ $=4 \mathrm{~V}$ We know, $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \mathrm{L} =\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L} =\frac{4}{(4-0) / 1 \times 10^{-3}}$ $\mathrm{~L} =10^{-3} \mathrm{H}$ $\text { Or } \mathrm{L} =1 \mathrm{mH}$
COMEDK 2011
Electro Magnetic Induction
154854
A current of 2.5 A flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is
1 $0.5 \mathrm{~Wb}$
2 $12.5 \mathrm{~Wb}$
3 zero
4 $2 \mathrm{~Wb}$
Explanation:
B Given- \(\mathrm{I}=2.5 \mathrm{~A} . \quad \mathrm{L}=5 \mathrm{H} \text {. }\) Magnetic flux,\(\quad \phi=\) ? \(\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}\) \(\mathrm{d} \phi=\mathrm{L} \cdot \mathrm{dI}\) or \(\phi=\mathrm{LI}\) \(\phi=5 \times 2.5\) \(\phi=12.5 \mathrm{~Wb}\)
COMEDK 2013
Electro Magnetic Induction
154857
If a change in current of $0.01 \mathrm{~A}$ in one coil produces a change in magnetic flux of $2 \times 10^{-2}$ weber in another coil, then the mutual inductance between coils is
1 0
2 $0.5 \mathrm{H}$
3 $2 \mathrm{H}$
4 $3 \mathrm{H}$
Explanation:
C Given that, Change in current $\Delta \mathrm{I}=0.01 \mathrm{~A}$, Magnetic flux $\Delta \phi=2 \times 10^{-2} \mathrm{~Wb}$ $\mathrm{M}=\text { ? }$ We know, $\Delta \phi=\mathrm{MI}$ $\mathrm{M}=\frac{\Delta \phi}{\Delta \mathrm{I}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \mathrm{H}$
154852
The current in a coil changes from $1 \mathrm{~mA}$ to 5 $\mathrm{mA}$ in $4 \mathrm{~ms}$. If the coefficient of self-inductance of the coil is $10 \mathrm{mH}$, the magnitude of "selfinduced" emf is
154853
The current in a coil changes from $4 \mathrm{~A}$ to zero in $1 \mathrm{~ms}$. The average emf induced in the coil is $4 \mathrm{~V}$. The self inductance, in $\mathrm{mH}$, between the coil is
1 2.5
2 0.25
3 25
4 1
Explanation:
D Given that, $\mathrm{I}_1=4 \mathrm{~A}$ $\text { And } \mathrm{I}_2=0 \mathrm{~A}$ $\mathrm{t}=1 \mathrm{~m} \mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ $=4 \mathrm{~V}$ We know, $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \mathrm{L} =\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L} =\frac{4}{(4-0) / 1 \times 10^{-3}}$ $\mathrm{~L} =10^{-3} \mathrm{H}$ $\text { Or } \mathrm{L} =1 \mathrm{mH}$
COMEDK 2011
Electro Magnetic Induction
154854
A current of 2.5 A flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is
1 $0.5 \mathrm{~Wb}$
2 $12.5 \mathrm{~Wb}$
3 zero
4 $2 \mathrm{~Wb}$
Explanation:
B Given- \(\mathrm{I}=2.5 \mathrm{~A} . \quad \mathrm{L}=5 \mathrm{H} \text {. }\) Magnetic flux,\(\quad \phi=\) ? \(\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}\) \(\mathrm{d} \phi=\mathrm{L} \cdot \mathrm{dI}\) or \(\phi=\mathrm{LI}\) \(\phi=5 \times 2.5\) \(\phi=12.5 \mathrm{~Wb}\)
COMEDK 2013
Electro Magnetic Induction
154857
If a change in current of $0.01 \mathrm{~A}$ in one coil produces a change in magnetic flux of $2 \times 10^{-2}$ weber in another coil, then the mutual inductance between coils is
1 0
2 $0.5 \mathrm{H}$
3 $2 \mathrm{H}$
4 $3 \mathrm{H}$
Explanation:
C Given that, Change in current $\Delta \mathrm{I}=0.01 \mathrm{~A}$, Magnetic flux $\Delta \phi=2 \times 10^{-2} \mathrm{~Wb}$ $\mathrm{M}=\text { ? }$ We know, $\Delta \phi=\mathrm{MI}$ $\mathrm{M}=\frac{\Delta \phi}{\Delta \mathrm{I}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \mathrm{H}$
154852
The current in a coil changes from $1 \mathrm{~mA}$ to 5 $\mathrm{mA}$ in $4 \mathrm{~ms}$. If the coefficient of self-inductance of the coil is $10 \mathrm{mH}$, the magnitude of "selfinduced" emf is
154853
The current in a coil changes from $4 \mathrm{~A}$ to zero in $1 \mathrm{~ms}$. The average emf induced in the coil is $4 \mathrm{~V}$. The self inductance, in $\mathrm{mH}$, between the coil is
1 2.5
2 0.25
3 25
4 1
Explanation:
D Given that, $\mathrm{I}_1=4 \mathrm{~A}$ $\text { And } \mathrm{I}_2=0 \mathrm{~A}$ $\mathrm{t}=1 \mathrm{~m} \mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ $=4 \mathrm{~V}$ We know, $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \mathrm{L} =\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L} =\frac{4}{(4-0) / 1 \times 10^{-3}}$ $\mathrm{~L} =10^{-3} \mathrm{H}$ $\text { Or } \mathrm{L} =1 \mathrm{mH}$
COMEDK 2011
Electro Magnetic Induction
154854
A current of 2.5 A flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is
1 $0.5 \mathrm{~Wb}$
2 $12.5 \mathrm{~Wb}$
3 zero
4 $2 \mathrm{~Wb}$
Explanation:
B Given- \(\mathrm{I}=2.5 \mathrm{~A} . \quad \mathrm{L}=5 \mathrm{H} \text {. }\) Magnetic flux,\(\quad \phi=\) ? \(\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}\) \(\mathrm{d} \phi=\mathrm{L} \cdot \mathrm{dI}\) or \(\phi=\mathrm{LI}\) \(\phi=5 \times 2.5\) \(\phi=12.5 \mathrm{~Wb}\)
COMEDK 2013
Electro Magnetic Induction
154857
If a change in current of $0.01 \mathrm{~A}$ in one coil produces a change in magnetic flux of $2 \times 10^{-2}$ weber in another coil, then the mutual inductance between coils is
1 0
2 $0.5 \mathrm{H}$
3 $2 \mathrm{H}$
4 $3 \mathrm{H}$
Explanation:
C Given that, Change in current $\Delta \mathrm{I}=0.01 \mathrm{~A}$, Magnetic flux $\Delta \phi=2 \times 10^{-2} \mathrm{~Wb}$ $\mathrm{M}=\text { ? }$ We know, $\Delta \phi=\mathrm{MI}$ $\mathrm{M}=\frac{\Delta \phi}{\Delta \mathrm{I}}=\frac{2 \times 10^{-2}}{0.01}$ $\mathrm{M}=2 \mathrm{H}$