154815
Two coils have mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $I=I_{0} \sin \omega t$, where $I_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
1 $12 \pi$
2 $8 \pi$
3 $5 \pi$
4 $2 \pi$
Explanation:
C Given that, $\mathrm{M}=0.005 \mathrm{H}, \mathrm{I}_{\mathrm{o}}=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ $I=I_{0} \sin \omega t$ We know that, Induced emf $(\varepsilon)=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=M \frac{d\left(I_{\circ} \sin \omega t\right)}{d t}$ $\varepsilon=\mathrm{MI}_{\mathrm{o}} \cos (\omega \mathrm{t}) . \omega$ $\varepsilon_{\max }=0.005 \times 10 \times 1 \times 100 \pi$ $\{\text { for maximum emf }=\cos (\omega \mathrm{t})=1 .\mathrm{at}, \mathrm{t}=0$ $\varepsilon_{\max }=5 \pi \mathrm{V}$
Manipal UGET-2015
Electro Magnetic Induction
154816
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R$ $\Omega$. This combination is moved in time $t$ second from a magnetic field $w_{1}$ Weber to $w_{2}$ Weber. The induced current in the circuit is
B We know that, $\text { Current, } \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}$ Also, $\operatorname{emf}(\varepsilon)=-\mathrm{n} \frac{\mathrm{d} \phi}{\mathrm{dt}}$ We get, $I=\frac{-n}{R} \frac{d \phi}{d t}$ Here, $\mathrm{d} \phi=\mathrm{w}_{2}-\mathrm{w}_{1}$ and $\mathrm{dt}=\mathrm{t}$ [Where, $\mathrm{w}_{1}$ and $\mathrm{w}_{2}$ are flux] $\therefore \quad \mathrm{I}=\frac{-\mathrm{n}}{\mathrm{R}}\left[\frac{\mathrm{w}_{2}-\mathrm{w}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}\right]$ Total resistance, $R^{\prime}=R+4 R=5 R$ $\mathrm{I}=\frac{-\mathrm{n}}{5 \mathrm{R}} \frac{\left[\mathrm{w}_{2}-\mathrm{w}_{1}\right]}{\mathrm{t}}$ $\left(t_{2}-t_{1}=t\right)$
Manipal UGET-2015
Electro Magnetic Induction
154819
Current in a coil changes from $5 \mathrm{~A}$ to $10 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If the coefficient of self-induction is $10 \mathrm{H}$, then the induced emf is
1 $112 \mathrm{~V}$
2 $250 \mathrm{~V}$
3 $125 \mathrm{~V}$
4 $230 \mathrm{~V}$
Explanation:
B Given that, Current $(\mathrm{dI})=10-5=5 \mathrm{~A}$ Total time $(\mathrm{dt})=0.2 \mathrm{sec}$ Self - inductance $(\mathrm{L})=10 \mathrm{H}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=10 \times \frac{5}{0.2}$ $\varepsilon=250 \mathrm{~V}$
Manipal UGET-2010
Electro Magnetic Induction
154820
In a coil when current changes from $10 \mathrm{~A}$ to $2 \mathrm{~A}$ in time $0.1 \mathrm{~s}$, induced emf is $3.28 \mathrm{~V}$. What is self-inductance of coil?
1 $4 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.04 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
C Given that, Current $(\mathrm{dI})=10-2=8 \mathrm{~A}$ Total time $(\mathrm{dt})=0.1 \mathrm{sec}$ Induced emf $(\varepsilon)=3.28 \mathrm{~V}$ We know that, $\mathrm{L}=\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L}=\frac{3.28}{8 / 0.1}$ $\mathrm{~L}=0.04 \mathrm{H}$
Manipal UGET-2010
Electro Magnetic Induction
154821
Two long straight conductors with currents $I_{1}$ and $I_{2}$ are placed along $X$-axis and $Y$-axis as shown in figure. The equation of locus of zero magnetic induction is
C If $\mathrm{G}(\mathrm{x}, \mathrm{y})$ is the point where the magnetic field is zero because direction of magnetic fields are in opposite direction. $\frac{\mu_{\circ}}{4 \pi} \cdot \frac{2 I_{1}}{y}=\frac{\mu_{\circ}}{4 \pi} \frac{2 I_{2}}{x}$ $y=\left(\frac{I_{1}}{I_{2}}\right) x$
154815
Two coils have mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $I=I_{0} \sin \omega t$, where $I_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
1 $12 \pi$
2 $8 \pi$
3 $5 \pi$
4 $2 \pi$
Explanation:
C Given that, $\mathrm{M}=0.005 \mathrm{H}, \mathrm{I}_{\mathrm{o}}=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ $I=I_{0} \sin \omega t$ We know that, Induced emf $(\varepsilon)=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=M \frac{d\left(I_{\circ} \sin \omega t\right)}{d t}$ $\varepsilon=\mathrm{MI}_{\mathrm{o}} \cos (\omega \mathrm{t}) . \omega$ $\varepsilon_{\max }=0.005 \times 10 \times 1 \times 100 \pi$ $\{\text { for maximum emf }=\cos (\omega \mathrm{t})=1 .\mathrm{at}, \mathrm{t}=0$ $\varepsilon_{\max }=5 \pi \mathrm{V}$
Manipal UGET-2015
Electro Magnetic Induction
154816
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R$ $\Omega$. This combination is moved in time $t$ second from a magnetic field $w_{1}$ Weber to $w_{2}$ Weber. The induced current in the circuit is
B We know that, $\text { Current, } \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}$ Also, $\operatorname{emf}(\varepsilon)=-\mathrm{n} \frac{\mathrm{d} \phi}{\mathrm{dt}}$ We get, $I=\frac{-n}{R} \frac{d \phi}{d t}$ Here, $\mathrm{d} \phi=\mathrm{w}_{2}-\mathrm{w}_{1}$ and $\mathrm{dt}=\mathrm{t}$ [Where, $\mathrm{w}_{1}$ and $\mathrm{w}_{2}$ are flux] $\therefore \quad \mathrm{I}=\frac{-\mathrm{n}}{\mathrm{R}}\left[\frac{\mathrm{w}_{2}-\mathrm{w}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}\right]$ Total resistance, $R^{\prime}=R+4 R=5 R$ $\mathrm{I}=\frac{-\mathrm{n}}{5 \mathrm{R}} \frac{\left[\mathrm{w}_{2}-\mathrm{w}_{1}\right]}{\mathrm{t}}$ $\left(t_{2}-t_{1}=t\right)$
Manipal UGET-2015
Electro Magnetic Induction
154819
Current in a coil changes from $5 \mathrm{~A}$ to $10 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If the coefficient of self-induction is $10 \mathrm{H}$, then the induced emf is
1 $112 \mathrm{~V}$
2 $250 \mathrm{~V}$
3 $125 \mathrm{~V}$
4 $230 \mathrm{~V}$
Explanation:
B Given that, Current $(\mathrm{dI})=10-5=5 \mathrm{~A}$ Total time $(\mathrm{dt})=0.2 \mathrm{sec}$ Self - inductance $(\mathrm{L})=10 \mathrm{H}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=10 \times \frac{5}{0.2}$ $\varepsilon=250 \mathrm{~V}$
Manipal UGET-2010
Electro Magnetic Induction
154820
In a coil when current changes from $10 \mathrm{~A}$ to $2 \mathrm{~A}$ in time $0.1 \mathrm{~s}$, induced emf is $3.28 \mathrm{~V}$. What is self-inductance of coil?
1 $4 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.04 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
C Given that, Current $(\mathrm{dI})=10-2=8 \mathrm{~A}$ Total time $(\mathrm{dt})=0.1 \mathrm{sec}$ Induced emf $(\varepsilon)=3.28 \mathrm{~V}$ We know that, $\mathrm{L}=\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L}=\frac{3.28}{8 / 0.1}$ $\mathrm{~L}=0.04 \mathrm{H}$
Manipal UGET-2010
Electro Magnetic Induction
154821
Two long straight conductors with currents $I_{1}$ and $I_{2}$ are placed along $X$-axis and $Y$-axis as shown in figure. The equation of locus of zero magnetic induction is
C If $\mathrm{G}(\mathrm{x}, \mathrm{y})$ is the point where the magnetic field is zero because direction of magnetic fields are in opposite direction. $\frac{\mu_{\circ}}{4 \pi} \cdot \frac{2 I_{1}}{y}=\frac{\mu_{\circ}}{4 \pi} \frac{2 I_{2}}{x}$ $y=\left(\frac{I_{1}}{I_{2}}\right) x$
154815
Two coils have mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $I=I_{0} \sin \omega t$, where $I_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
1 $12 \pi$
2 $8 \pi$
3 $5 \pi$
4 $2 \pi$
Explanation:
C Given that, $\mathrm{M}=0.005 \mathrm{H}, \mathrm{I}_{\mathrm{o}}=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ $I=I_{0} \sin \omega t$ We know that, Induced emf $(\varepsilon)=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=M \frac{d\left(I_{\circ} \sin \omega t\right)}{d t}$ $\varepsilon=\mathrm{MI}_{\mathrm{o}} \cos (\omega \mathrm{t}) . \omega$ $\varepsilon_{\max }=0.005 \times 10 \times 1 \times 100 \pi$ $\{\text { for maximum emf }=\cos (\omega \mathrm{t})=1 .\mathrm{at}, \mathrm{t}=0$ $\varepsilon_{\max }=5 \pi \mathrm{V}$
Manipal UGET-2015
Electro Magnetic Induction
154816
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R$ $\Omega$. This combination is moved in time $t$ second from a magnetic field $w_{1}$ Weber to $w_{2}$ Weber. The induced current in the circuit is
B We know that, $\text { Current, } \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}$ Also, $\operatorname{emf}(\varepsilon)=-\mathrm{n} \frac{\mathrm{d} \phi}{\mathrm{dt}}$ We get, $I=\frac{-n}{R} \frac{d \phi}{d t}$ Here, $\mathrm{d} \phi=\mathrm{w}_{2}-\mathrm{w}_{1}$ and $\mathrm{dt}=\mathrm{t}$ [Where, $\mathrm{w}_{1}$ and $\mathrm{w}_{2}$ are flux] $\therefore \quad \mathrm{I}=\frac{-\mathrm{n}}{\mathrm{R}}\left[\frac{\mathrm{w}_{2}-\mathrm{w}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}\right]$ Total resistance, $R^{\prime}=R+4 R=5 R$ $\mathrm{I}=\frac{-\mathrm{n}}{5 \mathrm{R}} \frac{\left[\mathrm{w}_{2}-\mathrm{w}_{1}\right]}{\mathrm{t}}$ $\left(t_{2}-t_{1}=t\right)$
Manipal UGET-2015
Electro Magnetic Induction
154819
Current in a coil changes from $5 \mathrm{~A}$ to $10 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If the coefficient of self-induction is $10 \mathrm{H}$, then the induced emf is
1 $112 \mathrm{~V}$
2 $250 \mathrm{~V}$
3 $125 \mathrm{~V}$
4 $230 \mathrm{~V}$
Explanation:
B Given that, Current $(\mathrm{dI})=10-5=5 \mathrm{~A}$ Total time $(\mathrm{dt})=0.2 \mathrm{sec}$ Self - inductance $(\mathrm{L})=10 \mathrm{H}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=10 \times \frac{5}{0.2}$ $\varepsilon=250 \mathrm{~V}$
Manipal UGET-2010
Electro Magnetic Induction
154820
In a coil when current changes from $10 \mathrm{~A}$ to $2 \mathrm{~A}$ in time $0.1 \mathrm{~s}$, induced emf is $3.28 \mathrm{~V}$. What is self-inductance of coil?
1 $4 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.04 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
C Given that, Current $(\mathrm{dI})=10-2=8 \mathrm{~A}$ Total time $(\mathrm{dt})=0.1 \mathrm{sec}$ Induced emf $(\varepsilon)=3.28 \mathrm{~V}$ We know that, $\mathrm{L}=\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L}=\frac{3.28}{8 / 0.1}$ $\mathrm{~L}=0.04 \mathrm{H}$
Manipal UGET-2010
Electro Magnetic Induction
154821
Two long straight conductors with currents $I_{1}$ and $I_{2}$ are placed along $X$-axis and $Y$-axis as shown in figure. The equation of locus of zero magnetic induction is
C If $\mathrm{G}(\mathrm{x}, \mathrm{y})$ is the point where the magnetic field is zero because direction of magnetic fields are in opposite direction. $\frac{\mu_{\circ}}{4 \pi} \cdot \frac{2 I_{1}}{y}=\frac{\mu_{\circ}}{4 \pi} \frac{2 I_{2}}{x}$ $y=\left(\frac{I_{1}}{I_{2}}\right) x$
154815
Two coils have mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $I=I_{0} \sin \omega t$, where $I_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
1 $12 \pi$
2 $8 \pi$
3 $5 \pi$
4 $2 \pi$
Explanation:
C Given that, $\mathrm{M}=0.005 \mathrm{H}, \mathrm{I}_{\mathrm{o}}=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ $I=I_{0} \sin \omega t$ We know that, Induced emf $(\varepsilon)=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=M \frac{d\left(I_{\circ} \sin \omega t\right)}{d t}$ $\varepsilon=\mathrm{MI}_{\mathrm{o}} \cos (\omega \mathrm{t}) . \omega$ $\varepsilon_{\max }=0.005 \times 10 \times 1 \times 100 \pi$ $\{\text { for maximum emf }=\cos (\omega \mathrm{t})=1 .\mathrm{at}, \mathrm{t}=0$ $\varepsilon_{\max }=5 \pi \mathrm{V}$
Manipal UGET-2015
Electro Magnetic Induction
154816
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R$ $\Omega$. This combination is moved in time $t$ second from a magnetic field $w_{1}$ Weber to $w_{2}$ Weber. The induced current in the circuit is
B We know that, $\text { Current, } \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}$ Also, $\operatorname{emf}(\varepsilon)=-\mathrm{n} \frac{\mathrm{d} \phi}{\mathrm{dt}}$ We get, $I=\frac{-n}{R} \frac{d \phi}{d t}$ Here, $\mathrm{d} \phi=\mathrm{w}_{2}-\mathrm{w}_{1}$ and $\mathrm{dt}=\mathrm{t}$ [Where, $\mathrm{w}_{1}$ and $\mathrm{w}_{2}$ are flux] $\therefore \quad \mathrm{I}=\frac{-\mathrm{n}}{\mathrm{R}}\left[\frac{\mathrm{w}_{2}-\mathrm{w}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}\right]$ Total resistance, $R^{\prime}=R+4 R=5 R$ $\mathrm{I}=\frac{-\mathrm{n}}{5 \mathrm{R}} \frac{\left[\mathrm{w}_{2}-\mathrm{w}_{1}\right]}{\mathrm{t}}$ $\left(t_{2}-t_{1}=t\right)$
Manipal UGET-2015
Electro Magnetic Induction
154819
Current in a coil changes from $5 \mathrm{~A}$ to $10 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If the coefficient of self-induction is $10 \mathrm{H}$, then the induced emf is
1 $112 \mathrm{~V}$
2 $250 \mathrm{~V}$
3 $125 \mathrm{~V}$
4 $230 \mathrm{~V}$
Explanation:
B Given that, Current $(\mathrm{dI})=10-5=5 \mathrm{~A}$ Total time $(\mathrm{dt})=0.2 \mathrm{sec}$ Self - inductance $(\mathrm{L})=10 \mathrm{H}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=10 \times \frac{5}{0.2}$ $\varepsilon=250 \mathrm{~V}$
Manipal UGET-2010
Electro Magnetic Induction
154820
In a coil when current changes from $10 \mathrm{~A}$ to $2 \mathrm{~A}$ in time $0.1 \mathrm{~s}$, induced emf is $3.28 \mathrm{~V}$. What is self-inductance of coil?
1 $4 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.04 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
C Given that, Current $(\mathrm{dI})=10-2=8 \mathrm{~A}$ Total time $(\mathrm{dt})=0.1 \mathrm{sec}$ Induced emf $(\varepsilon)=3.28 \mathrm{~V}$ We know that, $\mathrm{L}=\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L}=\frac{3.28}{8 / 0.1}$ $\mathrm{~L}=0.04 \mathrm{H}$
Manipal UGET-2010
Electro Magnetic Induction
154821
Two long straight conductors with currents $I_{1}$ and $I_{2}$ are placed along $X$-axis and $Y$-axis as shown in figure. The equation of locus of zero magnetic induction is
C If $\mathrm{G}(\mathrm{x}, \mathrm{y})$ is the point where the magnetic field is zero because direction of magnetic fields are in opposite direction. $\frac{\mu_{\circ}}{4 \pi} \cdot \frac{2 I_{1}}{y}=\frac{\mu_{\circ}}{4 \pi} \frac{2 I_{2}}{x}$ $y=\left(\frac{I_{1}}{I_{2}}\right) x$
154815
Two coils have mutual inductance $0.005 \mathrm{H}$. The current changes in the first coil according to equation $I=I_{0} \sin \omega t$, where $I_{0}=10 \mathrm{~A}$ and $\omega=$ $100 \pi \mathrm{rad} / \mathrm{s}$. The maximum value of emf in the second coil is
1 $12 \pi$
2 $8 \pi$
3 $5 \pi$
4 $2 \pi$
Explanation:
C Given that, $\mathrm{M}=0.005 \mathrm{H}, \mathrm{I}_{\mathrm{o}}=10 \mathrm{~A}$ and $\omega=100 \pi \mathrm{rad} / \mathrm{s}$ $I=I_{0} \sin \omega t$ We know that, Induced emf $(\varepsilon)=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=M \frac{d\left(I_{\circ} \sin \omega t\right)}{d t}$ $\varepsilon=\mathrm{MI}_{\mathrm{o}} \cos (\omega \mathrm{t}) . \omega$ $\varepsilon_{\max }=0.005 \times 10 \times 1 \times 100 \pi$ $\{\text { for maximum emf }=\cos (\omega \mathrm{t})=1 .\mathrm{at}, \mathrm{t}=0$ $\varepsilon_{\max }=5 \pi \mathrm{V}$
Manipal UGET-2015
Electro Magnetic Induction
154816
A coil having $n$ turns and resistance $R \Omega$ is connected with a galvanometer of resistance $4 R$ $\Omega$. This combination is moved in time $t$ second from a magnetic field $w_{1}$ Weber to $w_{2}$ Weber. The induced current in the circuit is
B We know that, $\text { Current, } \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}$ Also, $\operatorname{emf}(\varepsilon)=-\mathrm{n} \frac{\mathrm{d} \phi}{\mathrm{dt}}$ We get, $I=\frac{-n}{R} \frac{d \phi}{d t}$ Here, $\mathrm{d} \phi=\mathrm{w}_{2}-\mathrm{w}_{1}$ and $\mathrm{dt}=\mathrm{t}$ [Where, $\mathrm{w}_{1}$ and $\mathrm{w}_{2}$ are flux] $\therefore \quad \mathrm{I}=\frac{-\mathrm{n}}{\mathrm{R}}\left[\frac{\mathrm{w}_{2}-\mathrm{w}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}\right]$ Total resistance, $R^{\prime}=R+4 R=5 R$ $\mathrm{I}=\frac{-\mathrm{n}}{5 \mathrm{R}} \frac{\left[\mathrm{w}_{2}-\mathrm{w}_{1}\right]}{\mathrm{t}}$ $\left(t_{2}-t_{1}=t\right)$
Manipal UGET-2015
Electro Magnetic Induction
154819
Current in a coil changes from $5 \mathrm{~A}$ to $10 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If the coefficient of self-induction is $10 \mathrm{H}$, then the induced emf is
1 $112 \mathrm{~V}$
2 $250 \mathrm{~V}$
3 $125 \mathrm{~V}$
4 $230 \mathrm{~V}$
Explanation:
B Given that, Current $(\mathrm{dI})=10-5=5 \mathrm{~A}$ Total time $(\mathrm{dt})=0.2 \mathrm{sec}$ Self - inductance $(\mathrm{L})=10 \mathrm{H}$ We know that, Induced emf $(\varepsilon)=\mathrm{L} \cdot \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=10 \times \frac{5}{0.2}$ $\varepsilon=250 \mathrm{~V}$
Manipal UGET-2010
Electro Magnetic Induction
154820
In a coil when current changes from $10 \mathrm{~A}$ to $2 \mathrm{~A}$ in time $0.1 \mathrm{~s}$, induced emf is $3.28 \mathrm{~V}$. What is self-inductance of coil?
1 $4 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.04 \mathrm{H}$
4 $5 \mathrm{H}$
Explanation:
C Given that, Current $(\mathrm{dI})=10-2=8 \mathrm{~A}$ Total time $(\mathrm{dt})=0.1 \mathrm{sec}$ Induced emf $(\varepsilon)=3.28 \mathrm{~V}$ We know that, $\mathrm{L}=\frac{\varepsilon}{\mathrm{dI} / \mathrm{dt}}$ $\mathrm{L}=\frac{3.28}{8 / 0.1}$ $\mathrm{~L}=0.04 \mathrm{H}$
Manipal UGET-2010
Electro Magnetic Induction
154821
Two long straight conductors with currents $I_{1}$ and $I_{2}$ are placed along $X$-axis and $Y$-axis as shown in figure. The equation of locus of zero magnetic induction is
C If $\mathrm{G}(\mathrm{x}, \mathrm{y})$ is the point where the magnetic field is zero because direction of magnetic fields are in opposite direction. $\frac{\mu_{\circ}}{4 \pi} \cdot \frac{2 I_{1}}{y}=\frac{\mu_{\circ}}{4 \pi} \frac{2 I_{2}}{x}$ $y=\left(\frac{I_{1}}{I_{2}}\right) x$
