154903
If the current through a coil changes from $1 \mathrm{~A}$ to $3 \mathrm{~A}$ in $0.02 \mathrm{~s}$ to produce an emf of $6 \mathrm{~V}$, then the self-inductance of the coil is
1 $0.12 \mathrm{H}$
2 $0.06 \mathrm{H}$
3 $0.02 \mathrm{H}$
4 $0.01 \mathrm{H}$
Explanation:
B Given, Change in current $\Delta \mathrm{i}=(3-1)=2 \mathrm{~A}$ Change in time $\Delta \mathrm{t}=0.02 \mathrm{sec}$ Self induced emf $\varepsilon=6 \mathrm{~V}$ $\therefore \varepsilon=\mathrm{L} \frac{\Delta \mathrm{i}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{i}}=\frac{6 \times 0.02}{2}=3 \times 0.02=0.06 \mathrm{H}$
J and K CET-2016
Electro Magnetic Induction
154904
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent selfinductance $L$ for the assembly is approximately
1 $50 \mathrm{mH}$
2 $36 \mathrm{mH}$
3 $28 \mathrm{mH}$
4 $18 \mathrm{mH}$
Explanation:
C Given Self inductance $\mathrm{L}_{1}=6 \mathrm{mH}$ $\mathrm{L}_{2}=8 \mathrm{mH}$ For highest co-efficient of coupling $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+2 \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\therefore \mathrm{L}_{\mathrm{eq}}=6+8+2 \sqrt{6 \times 8}$ $\mathrm{~L}_{\mathrm{eq}}=14+2 \sqrt{48}=27.85$ $\mathrm{~L}_{\mathrm{eq}} \approx 28 \mathrm{mH}$
WB JEE 2016
Electro Magnetic Induction
154905
In an inductor of self-inductance $L=2 \mathrm{mH}$, current changes with time according to the relation $I=t^{2} e^{-t}$. At what time emf is zero ?
1 $3 \mathrm{~s}$
2 $4 \mathrm{~s}$
3 I s
4 $2 \mathrm{~s}$
Explanation:
D Given Self inductance $\mathrm{L}=2 \mathrm{mH}$ Current $\mathrm{I}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ $\because$ Induced emf $\varepsilon=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\varepsilon=-\mathrm{L} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right)$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)\right]$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]$ For emf equal to zero $-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]=0$ $\Rightarrow \mathrm{te}^{-\mathrm{t}}[2-\mathrm{t}]=0 \Rightarrow \mathrm{t}=2 \mathrm{sec}$
UP CPMT-2009
Electro Magnetic Induction
154906
A wire in the form of a square of side a carries a current $i$. Then the magnetic induction at the centre of the square wire is $\left(\right.$ magnetic permeability of free space $\left.=\mu_{0}\right)$.
C Magnetic field due to side $\mathrm{AB}$ at point $\mathrm{P}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left[\sin \alpha_{1}+\sin \alpha_{2}\right]$ $\alpha_{1}=45^{\circ}, \alpha_{2}=45^{\circ}$ and $\mathrm{r}=\mathrm{a} / 2$ $\therefore \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a} / 2}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}} \frac{2}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}$ Total magnetic field at point $\mathrm{P}$ $\therefore \mathrm{B}_{\text {total }}=4 \times \mathrm{B}_{1}$ $=4 \frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}=\frac{4 \mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $\mathrm{B}_{\text {total }}=\frac{2 \sqrt{2} \mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
UP CPMT-2014
Electro Magnetic Induction
154907
Two coils have the mutual inductance $0.05 \mathrm{H}$. The current changes in the first coil as $I=I_{0}$ $\sin \omega t$, where $I_{0}=1 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum emf induced in secondary coil is
154903
If the current through a coil changes from $1 \mathrm{~A}$ to $3 \mathrm{~A}$ in $0.02 \mathrm{~s}$ to produce an emf of $6 \mathrm{~V}$, then the self-inductance of the coil is
1 $0.12 \mathrm{H}$
2 $0.06 \mathrm{H}$
3 $0.02 \mathrm{H}$
4 $0.01 \mathrm{H}$
Explanation:
B Given, Change in current $\Delta \mathrm{i}=(3-1)=2 \mathrm{~A}$ Change in time $\Delta \mathrm{t}=0.02 \mathrm{sec}$ Self induced emf $\varepsilon=6 \mathrm{~V}$ $\therefore \varepsilon=\mathrm{L} \frac{\Delta \mathrm{i}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{i}}=\frac{6 \times 0.02}{2}=3 \times 0.02=0.06 \mathrm{H}$
J and K CET-2016
Electro Magnetic Induction
154904
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent selfinductance $L$ for the assembly is approximately
1 $50 \mathrm{mH}$
2 $36 \mathrm{mH}$
3 $28 \mathrm{mH}$
4 $18 \mathrm{mH}$
Explanation:
C Given Self inductance $\mathrm{L}_{1}=6 \mathrm{mH}$ $\mathrm{L}_{2}=8 \mathrm{mH}$ For highest co-efficient of coupling $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+2 \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\therefore \mathrm{L}_{\mathrm{eq}}=6+8+2 \sqrt{6 \times 8}$ $\mathrm{~L}_{\mathrm{eq}}=14+2 \sqrt{48}=27.85$ $\mathrm{~L}_{\mathrm{eq}} \approx 28 \mathrm{mH}$
WB JEE 2016
Electro Magnetic Induction
154905
In an inductor of self-inductance $L=2 \mathrm{mH}$, current changes with time according to the relation $I=t^{2} e^{-t}$. At what time emf is zero ?
1 $3 \mathrm{~s}$
2 $4 \mathrm{~s}$
3 I s
4 $2 \mathrm{~s}$
Explanation:
D Given Self inductance $\mathrm{L}=2 \mathrm{mH}$ Current $\mathrm{I}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ $\because$ Induced emf $\varepsilon=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\varepsilon=-\mathrm{L} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right)$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)\right]$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]$ For emf equal to zero $-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]=0$ $\Rightarrow \mathrm{te}^{-\mathrm{t}}[2-\mathrm{t}]=0 \Rightarrow \mathrm{t}=2 \mathrm{sec}$
UP CPMT-2009
Electro Magnetic Induction
154906
A wire in the form of a square of side a carries a current $i$. Then the magnetic induction at the centre of the square wire is $\left(\right.$ magnetic permeability of free space $\left.=\mu_{0}\right)$.
C Magnetic field due to side $\mathrm{AB}$ at point $\mathrm{P}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left[\sin \alpha_{1}+\sin \alpha_{2}\right]$ $\alpha_{1}=45^{\circ}, \alpha_{2}=45^{\circ}$ and $\mathrm{r}=\mathrm{a} / 2$ $\therefore \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a} / 2}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}} \frac{2}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}$ Total magnetic field at point $\mathrm{P}$ $\therefore \mathrm{B}_{\text {total }}=4 \times \mathrm{B}_{1}$ $=4 \frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}=\frac{4 \mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $\mathrm{B}_{\text {total }}=\frac{2 \sqrt{2} \mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
UP CPMT-2014
Electro Magnetic Induction
154907
Two coils have the mutual inductance $0.05 \mathrm{H}$. The current changes in the first coil as $I=I_{0}$ $\sin \omega t$, where $I_{0}=1 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum emf induced in secondary coil is
154903
If the current through a coil changes from $1 \mathrm{~A}$ to $3 \mathrm{~A}$ in $0.02 \mathrm{~s}$ to produce an emf of $6 \mathrm{~V}$, then the self-inductance of the coil is
1 $0.12 \mathrm{H}$
2 $0.06 \mathrm{H}$
3 $0.02 \mathrm{H}$
4 $0.01 \mathrm{H}$
Explanation:
B Given, Change in current $\Delta \mathrm{i}=(3-1)=2 \mathrm{~A}$ Change in time $\Delta \mathrm{t}=0.02 \mathrm{sec}$ Self induced emf $\varepsilon=6 \mathrm{~V}$ $\therefore \varepsilon=\mathrm{L} \frac{\Delta \mathrm{i}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{i}}=\frac{6 \times 0.02}{2}=3 \times 0.02=0.06 \mathrm{H}$
J and K CET-2016
Electro Magnetic Induction
154904
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent selfinductance $L$ for the assembly is approximately
1 $50 \mathrm{mH}$
2 $36 \mathrm{mH}$
3 $28 \mathrm{mH}$
4 $18 \mathrm{mH}$
Explanation:
C Given Self inductance $\mathrm{L}_{1}=6 \mathrm{mH}$ $\mathrm{L}_{2}=8 \mathrm{mH}$ For highest co-efficient of coupling $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+2 \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\therefore \mathrm{L}_{\mathrm{eq}}=6+8+2 \sqrt{6 \times 8}$ $\mathrm{~L}_{\mathrm{eq}}=14+2 \sqrt{48}=27.85$ $\mathrm{~L}_{\mathrm{eq}} \approx 28 \mathrm{mH}$
WB JEE 2016
Electro Magnetic Induction
154905
In an inductor of self-inductance $L=2 \mathrm{mH}$, current changes with time according to the relation $I=t^{2} e^{-t}$. At what time emf is zero ?
1 $3 \mathrm{~s}$
2 $4 \mathrm{~s}$
3 I s
4 $2 \mathrm{~s}$
Explanation:
D Given Self inductance $\mathrm{L}=2 \mathrm{mH}$ Current $\mathrm{I}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ $\because$ Induced emf $\varepsilon=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\varepsilon=-\mathrm{L} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right)$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)\right]$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]$ For emf equal to zero $-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]=0$ $\Rightarrow \mathrm{te}^{-\mathrm{t}}[2-\mathrm{t}]=0 \Rightarrow \mathrm{t}=2 \mathrm{sec}$
UP CPMT-2009
Electro Magnetic Induction
154906
A wire in the form of a square of side a carries a current $i$. Then the magnetic induction at the centre of the square wire is $\left(\right.$ magnetic permeability of free space $\left.=\mu_{0}\right)$.
C Magnetic field due to side $\mathrm{AB}$ at point $\mathrm{P}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left[\sin \alpha_{1}+\sin \alpha_{2}\right]$ $\alpha_{1}=45^{\circ}, \alpha_{2}=45^{\circ}$ and $\mathrm{r}=\mathrm{a} / 2$ $\therefore \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a} / 2}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}} \frac{2}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}$ Total magnetic field at point $\mathrm{P}$ $\therefore \mathrm{B}_{\text {total }}=4 \times \mathrm{B}_{1}$ $=4 \frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}=\frac{4 \mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $\mathrm{B}_{\text {total }}=\frac{2 \sqrt{2} \mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
UP CPMT-2014
Electro Magnetic Induction
154907
Two coils have the mutual inductance $0.05 \mathrm{H}$. The current changes in the first coil as $I=I_{0}$ $\sin \omega t$, where $I_{0}=1 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum emf induced in secondary coil is
154903
If the current through a coil changes from $1 \mathrm{~A}$ to $3 \mathrm{~A}$ in $0.02 \mathrm{~s}$ to produce an emf of $6 \mathrm{~V}$, then the self-inductance of the coil is
1 $0.12 \mathrm{H}$
2 $0.06 \mathrm{H}$
3 $0.02 \mathrm{H}$
4 $0.01 \mathrm{H}$
Explanation:
B Given, Change in current $\Delta \mathrm{i}=(3-1)=2 \mathrm{~A}$ Change in time $\Delta \mathrm{t}=0.02 \mathrm{sec}$ Self induced emf $\varepsilon=6 \mathrm{~V}$ $\therefore \varepsilon=\mathrm{L} \frac{\Delta \mathrm{i}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{i}}=\frac{6 \times 0.02}{2}=3 \times 0.02=0.06 \mathrm{H}$
J and K CET-2016
Electro Magnetic Induction
154904
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent selfinductance $L$ for the assembly is approximately
1 $50 \mathrm{mH}$
2 $36 \mathrm{mH}$
3 $28 \mathrm{mH}$
4 $18 \mathrm{mH}$
Explanation:
C Given Self inductance $\mathrm{L}_{1}=6 \mathrm{mH}$ $\mathrm{L}_{2}=8 \mathrm{mH}$ For highest co-efficient of coupling $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+2 \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\therefore \mathrm{L}_{\mathrm{eq}}=6+8+2 \sqrt{6 \times 8}$ $\mathrm{~L}_{\mathrm{eq}}=14+2 \sqrt{48}=27.85$ $\mathrm{~L}_{\mathrm{eq}} \approx 28 \mathrm{mH}$
WB JEE 2016
Electro Magnetic Induction
154905
In an inductor of self-inductance $L=2 \mathrm{mH}$, current changes with time according to the relation $I=t^{2} e^{-t}$. At what time emf is zero ?
1 $3 \mathrm{~s}$
2 $4 \mathrm{~s}$
3 I s
4 $2 \mathrm{~s}$
Explanation:
D Given Self inductance $\mathrm{L}=2 \mathrm{mH}$ Current $\mathrm{I}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ $\because$ Induced emf $\varepsilon=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\varepsilon=-\mathrm{L} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right)$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)\right]$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]$ For emf equal to zero $-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]=0$ $\Rightarrow \mathrm{te}^{-\mathrm{t}}[2-\mathrm{t}]=0 \Rightarrow \mathrm{t}=2 \mathrm{sec}$
UP CPMT-2009
Electro Magnetic Induction
154906
A wire in the form of a square of side a carries a current $i$. Then the magnetic induction at the centre of the square wire is $\left(\right.$ magnetic permeability of free space $\left.=\mu_{0}\right)$.
C Magnetic field due to side $\mathrm{AB}$ at point $\mathrm{P}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left[\sin \alpha_{1}+\sin \alpha_{2}\right]$ $\alpha_{1}=45^{\circ}, \alpha_{2}=45^{\circ}$ and $\mathrm{r}=\mathrm{a} / 2$ $\therefore \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a} / 2}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}} \frac{2}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}$ Total magnetic field at point $\mathrm{P}$ $\therefore \mathrm{B}_{\text {total }}=4 \times \mathrm{B}_{1}$ $=4 \frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}=\frac{4 \mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $\mathrm{B}_{\text {total }}=\frac{2 \sqrt{2} \mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
UP CPMT-2014
Electro Magnetic Induction
154907
Two coils have the mutual inductance $0.05 \mathrm{H}$. The current changes in the first coil as $I=I_{0}$ $\sin \omega t$, where $I_{0}=1 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum emf induced in secondary coil is
154903
If the current through a coil changes from $1 \mathrm{~A}$ to $3 \mathrm{~A}$ in $0.02 \mathrm{~s}$ to produce an emf of $6 \mathrm{~V}$, then the self-inductance of the coil is
1 $0.12 \mathrm{H}$
2 $0.06 \mathrm{H}$
3 $0.02 \mathrm{H}$
4 $0.01 \mathrm{H}$
Explanation:
B Given, Change in current $\Delta \mathrm{i}=(3-1)=2 \mathrm{~A}$ Change in time $\Delta \mathrm{t}=0.02 \mathrm{sec}$ Self induced emf $\varepsilon=6 \mathrm{~V}$ $\therefore \varepsilon=\mathrm{L} \frac{\Delta \mathrm{i}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{i}}=\frac{6 \times 0.02}{2}=3 \times 0.02=0.06 \mathrm{H}$
J and K CET-2016
Electro Magnetic Induction
154904
Two coils of self-inductances $6 \mathrm{mH}$ and $8 \mathrm{mH}$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent selfinductance $L$ for the assembly is approximately
1 $50 \mathrm{mH}$
2 $36 \mathrm{mH}$
3 $28 \mathrm{mH}$
4 $18 \mathrm{mH}$
Explanation:
C Given Self inductance $\mathrm{L}_{1}=6 \mathrm{mH}$ $\mathrm{L}_{2}=8 \mathrm{mH}$ For highest co-efficient of coupling $\mathrm{L}_{\mathrm{eq}}=\mathrm{L}_{1}+\mathrm{L}_{2}+2 \sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}$ $\therefore \mathrm{L}_{\mathrm{eq}}=6+8+2 \sqrt{6 \times 8}$ $\mathrm{~L}_{\mathrm{eq}}=14+2 \sqrt{48}=27.85$ $\mathrm{~L}_{\mathrm{eq}} \approx 28 \mathrm{mH}$
WB JEE 2016
Electro Magnetic Induction
154905
In an inductor of self-inductance $L=2 \mathrm{mH}$, current changes with time according to the relation $I=t^{2} e^{-t}$. At what time emf is zero ?
1 $3 \mathrm{~s}$
2 $4 \mathrm{~s}$
3 I s
4 $2 \mathrm{~s}$
Explanation:
D Given Self inductance $\mathrm{L}=2 \mathrm{mH}$ Current $\mathrm{I}=\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}$ $\because$ Induced emf $\varepsilon=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$ $\varepsilon=-\mathrm{L} \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right)$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}+\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}(-1)\right]$ $\varepsilon=-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]$ For emf equal to zero $-\mathrm{L}\left[2 \mathrm{te}^{-\mathrm{t}}-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}\right]=0$ $\Rightarrow \mathrm{te}^{-\mathrm{t}}[2-\mathrm{t}]=0 \Rightarrow \mathrm{t}=2 \mathrm{sec}$
UP CPMT-2009
Electro Magnetic Induction
154906
A wire in the form of a square of side a carries a current $i$. Then the magnetic induction at the centre of the square wire is $\left(\right.$ magnetic permeability of free space $\left.=\mu_{0}\right)$.
C Magnetic field due to side $\mathrm{AB}$ at point $\mathrm{P}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}\left[\sin \alpha_{1}+\sin \alpha_{2}\right]$ $\alpha_{1}=45^{\circ}, \alpha_{2}=45^{\circ}$ and $\mathrm{r}=\mathrm{a} / 2$ $\therefore \mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{a} / 2}\left[\sin 45^{\circ}+\sin 45^{\circ}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}} \frac{2}{\sqrt{2}}=\frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}$ Total magnetic field at point $\mathrm{P}$ $\therefore \mathrm{B}_{\text {total }}=4 \times \mathrm{B}_{1}$ $=4 \frac{\mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}}=\frac{4 \mu_{0} \mathrm{i}}{\sqrt{2} \pi \mathrm{a}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $\mathrm{B}_{\text {total }}=\frac{2 \sqrt{2} \mu_{0} \mathrm{i}}{\pi \mathrm{a}}$
UP CPMT-2014
Electro Magnetic Induction
154907
Two coils have the mutual inductance $0.05 \mathrm{H}$. The current changes in the first coil as $I=I_{0}$ $\sin \omega t$, where $I_{0}=1 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$. The maximum emf induced in secondary coil is
