154570
The normal magnetic flux passing through a coil changes with time according to the equation $\phi=6 t^{2}-5 t+1$. What is the magnitude of the induced current at $t=0.253 \mathrm{~s}$ and resistance $10 \Omega$ ?
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $0.2 \mathrm{~A}$
Explanation:
D Given that, Resistance $(\mathrm{R})=10 \Omega$ Magnetic flux, $\phi=6 \mathrm{t}^{2}-5 \mathrm{t}+1$ Now, induced EMF- $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=12 \mathrm{t}-5$ At, time $(\mathrm{t})=0.253 \mathrm{sec}$ $|\varepsilon|=12 \times 0.253-5$ $\varepsilon=1.96$ $\therefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{1.96}{10}=0.196 \approx 0.2 \mathrm{Amp}$
COMEDK 2014
Electro Magnetic Induction
154571
The magnetic flux of $3 \times 10^{-4} \mathrm{~Wb}$ are passing through a coil of 100 turns. If the e.m.f. induced in the coil is $1.5 \mathrm{~V}$, the time interval will be
1 $1 \mathrm{sec}$
2 $0.1 \mathrm{sec}$
3 $0.02 \mathrm{sec}$
4 $0.4 \mathrm{sec}$
Explanation:
C Given that, Magnetic flux, $\phi=3 \times 10^{-4} \mathrm{~Wb}$ $\mathrm{N}=100 \text { turns }$ EMF, $\varepsilon=1.5 \mathrm{~V}$ We know that, $|\varepsilon| =\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\mathrm{dt} =\frac{\mathrm{Nd} \phi}{\varepsilon}$ $\mathrm{dt} =\frac{100 \times 3 \times 10^{-4}}{1.5}$ $\mathrm{dt} =2 \times 10^{-2} \mathrm{~s}$ $\mathrm{dt} =0.02 \mathrm{sec}$ Time interval will be $0.02 \mathrm{sec}$.
COMEDK 2016]
Electro Magnetic Induction
154572
Assertion: An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason: Efficiency of electric motor depends only on magnitude of back emf.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is
Explanation:
C Efficiency $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}}$ Output power is given by $\varepsilon \mathrm{I}=\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}$ To obtain the maximum power differentiating $\mathrm{eq}^{\mathrm{n}}$ (i) with respect $\mathrm{e}$ and equating to zero $\frac{\mathrm{d}}{\mathrm{d} \varepsilon}\left[\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}\right]=0$ $\frac{1}{\mathrm{R}}[\mathrm{E}-2 \varepsilon]=0$ $\mathrm{E}=2 \varepsilon$ $\varepsilon=\frac{\mathrm{E}}{2}$ Thus back emf $(\varepsilon)$ becomes equal to half of the applied voltage
AIIMS-2008
Electro Magnetic Induction
154573
The flux linked with a coil at any instant ' $t$ ' is given by $\phi=10 t^{2}-50 t+250$. The induced emf at $\mathbf{t}=\mathbf{3 s}$ is
1 $-190 \mathrm{~V}$
2 $-10 \mathrm{~V}$
3 $10 \mathrm{~V}$
4 $190 \mathrm{~V}$
Explanation:
B Given that, $\phi=10 \mathrm{t}^{2}-50 \mathrm{t}+250$ We know that, induced emf $\varepsilon= =\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon =\frac{-\mathrm{d}}{\mathrm{dt}}\left(10 \mathrm{t}^{2}-50 \mathrm{t}+250\right)$ $\varepsilon =-(20 \mathrm{t}-50)$ At, time $(\mathrm{t})=3 \mathrm{sec}$ then, $\varepsilon=-(20 \times 3-50)$ $\varepsilon=-60+50$ $\varepsilon=-10 \mathrm{~V}$
154570
The normal magnetic flux passing through a coil changes with time according to the equation $\phi=6 t^{2}-5 t+1$. What is the magnitude of the induced current at $t=0.253 \mathrm{~s}$ and resistance $10 \Omega$ ?
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $0.2 \mathrm{~A}$
Explanation:
D Given that, Resistance $(\mathrm{R})=10 \Omega$ Magnetic flux, $\phi=6 \mathrm{t}^{2}-5 \mathrm{t}+1$ Now, induced EMF- $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=12 \mathrm{t}-5$ At, time $(\mathrm{t})=0.253 \mathrm{sec}$ $|\varepsilon|=12 \times 0.253-5$ $\varepsilon=1.96$ $\therefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{1.96}{10}=0.196 \approx 0.2 \mathrm{Amp}$
COMEDK 2014
Electro Magnetic Induction
154571
The magnetic flux of $3 \times 10^{-4} \mathrm{~Wb}$ are passing through a coil of 100 turns. If the e.m.f. induced in the coil is $1.5 \mathrm{~V}$, the time interval will be
1 $1 \mathrm{sec}$
2 $0.1 \mathrm{sec}$
3 $0.02 \mathrm{sec}$
4 $0.4 \mathrm{sec}$
Explanation:
C Given that, Magnetic flux, $\phi=3 \times 10^{-4} \mathrm{~Wb}$ $\mathrm{N}=100 \text { turns }$ EMF, $\varepsilon=1.5 \mathrm{~V}$ We know that, $|\varepsilon| =\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\mathrm{dt} =\frac{\mathrm{Nd} \phi}{\varepsilon}$ $\mathrm{dt} =\frac{100 \times 3 \times 10^{-4}}{1.5}$ $\mathrm{dt} =2 \times 10^{-2} \mathrm{~s}$ $\mathrm{dt} =0.02 \mathrm{sec}$ Time interval will be $0.02 \mathrm{sec}$.
COMEDK 2016]
Electro Magnetic Induction
154572
Assertion: An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason: Efficiency of electric motor depends only on magnitude of back emf.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is
Explanation:
C Efficiency $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}}$ Output power is given by $\varepsilon \mathrm{I}=\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}$ To obtain the maximum power differentiating $\mathrm{eq}^{\mathrm{n}}$ (i) with respect $\mathrm{e}$ and equating to zero $\frac{\mathrm{d}}{\mathrm{d} \varepsilon}\left[\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}\right]=0$ $\frac{1}{\mathrm{R}}[\mathrm{E}-2 \varepsilon]=0$ $\mathrm{E}=2 \varepsilon$ $\varepsilon=\frac{\mathrm{E}}{2}$ Thus back emf $(\varepsilon)$ becomes equal to half of the applied voltage
AIIMS-2008
Electro Magnetic Induction
154573
The flux linked with a coil at any instant ' $t$ ' is given by $\phi=10 t^{2}-50 t+250$. The induced emf at $\mathbf{t}=\mathbf{3 s}$ is
1 $-190 \mathrm{~V}$
2 $-10 \mathrm{~V}$
3 $10 \mathrm{~V}$
4 $190 \mathrm{~V}$
Explanation:
B Given that, $\phi=10 \mathrm{t}^{2}-50 \mathrm{t}+250$ We know that, induced emf $\varepsilon= =\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon =\frac{-\mathrm{d}}{\mathrm{dt}}\left(10 \mathrm{t}^{2}-50 \mathrm{t}+250\right)$ $\varepsilon =-(20 \mathrm{t}-50)$ At, time $(\mathrm{t})=3 \mathrm{sec}$ then, $\varepsilon=-(20 \times 3-50)$ $\varepsilon=-60+50$ $\varepsilon=-10 \mathrm{~V}$
154570
The normal magnetic flux passing through a coil changes with time according to the equation $\phi=6 t^{2}-5 t+1$. What is the magnitude of the induced current at $t=0.253 \mathrm{~s}$ and resistance $10 \Omega$ ?
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $0.2 \mathrm{~A}$
Explanation:
D Given that, Resistance $(\mathrm{R})=10 \Omega$ Magnetic flux, $\phi=6 \mathrm{t}^{2}-5 \mathrm{t}+1$ Now, induced EMF- $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=12 \mathrm{t}-5$ At, time $(\mathrm{t})=0.253 \mathrm{sec}$ $|\varepsilon|=12 \times 0.253-5$ $\varepsilon=1.96$ $\therefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{1.96}{10}=0.196 \approx 0.2 \mathrm{Amp}$
COMEDK 2014
Electro Magnetic Induction
154571
The magnetic flux of $3 \times 10^{-4} \mathrm{~Wb}$ are passing through a coil of 100 turns. If the e.m.f. induced in the coil is $1.5 \mathrm{~V}$, the time interval will be
1 $1 \mathrm{sec}$
2 $0.1 \mathrm{sec}$
3 $0.02 \mathrm{sec}$
4 $0.4 \mathrm{sec}$
Explanation:
C Given that, Magnetic flux, $\phi=3 \times 10^{-4} \mathrm{~Wb}$ $\mathrm{N}=100 \text { turns }$ EMF, $\varepsilon=1.5 \mathrm{~V}$ We know that, $|\varepsilon| =\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\mathrm{dt} =\frac{\mathrm{Nd} \phi}{\varepsilon}$ $\mathrm{dt} =\frac{100 \times 3 \times 10^{-4}}{1.5}$ $\mathrm{dt} =2 \times 10^{-2} \mathrm{~s}$ $\mathrm{dt} =0.02 \mathrm{sec}$ Time interval will be $0.02 \mathrm{sec}$.
COMEDK 2016]
Electro Magnetic Induction
154572
Assertion: An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason: Efficiency of electric motor depends only on magnitude of back emf.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is
Explanation:
C Efficiency $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}}$ Output power is given by $\varepsilon \mathrm{I}=\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}$ To obtain the maximum power differentiating $\mathrm{eq}^{\mathrm{n}}$ (i) with respect $\mathrm{e}$ and equating to zero $\frac{\mathrm{d}}{\mathrm{d} \varepsilon}\left[\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}\right]=0$ $\frac{1}{\mathrm{R}}[\mathrm{E}-2 \varepsilon]=0$ $\mathrm{E}=2 \varepsilon$ $\varepsilon=\frac{\mathrm{E}}{2}$ Thus back emf $(\varepsilon)$ becomes equal to half of the applied voltage
AIIMS-2008
Electro Magnetic Induction
154573
The flux linked with a coil at any instant ' $t$ ' is given by $\phi=10 t^{2}-50 t+250$. The induced emf at $\mathbf{t}=\mathbf{3 s}$ is
1 $-190 \mathrm{~V}$
2 $-10 \mathrm{~V}$
3 $10 \mathrm{~V}$
4 $190 \mathrm{~V}$
Explanation:
B Given that, $\phi=10 \mathrm{t}^{2}-50 \mathrm{t}+250$ We know that, induced emf $\varepsilon= =\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon =\frac{-\mathrm{d}}{\mathrm{dt}}\left(10 \mathrm{t}^{2}-50 \mathrm{t}+250\right)$ $\varepsilon =-(20 \mathrm{t}-50)$ At, time $(\mathrm{t})=3 \mathrm{sec}$ then, $\varepsilon=-(20 \times 3-50)$ $\varepsilon=-60+50$ $\varepsilon=-10 \mathrm{~V}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154570
The normal magnetic flux passing through a coil changes with time according to the equation $\phi=6 t^{2}-5 t+1$. What is the magnitude of the induced current at $t=0.253 \mathrm{~s}$ and resistance $10 \Omega$ ?
1 $1.2 \mathrm{~A}$
2 $0.8 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $0.2 \mathrm{~A}$
Explanation:
D Given that, Resistance $(\mathrm{R})=10 \Omega$ Magnetic flux, $\phi=6 \mathrm{t}^{2}-5 \mathrm{t}+1$ Now, induced EMF- $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=12 \mathrm{t}-5$ At, time $(\mathrm{t})=0.253 \mathrm{sec}$ $|\varepsilon|=12 \times 0.253-5$ $\varepsilon=1.96$ $\therefore \quad \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{1.96}{10}=0.196 \approx 0.2 \mathrm{Amp}$
COMEDK 2014
Electro Magnetic Induction
154571
The magnetic flux of $3 \times 10^{-4} \mathrm{~Wb}$ are passing through a coil of 100 turns. If the e.m.f. induced in the coil is $1.5 \mathrm{~V}$, the time interval will be
1 $1 \mathrm{sec}$
2 $0.1 \mathrm{sec}$
3 $0.02 \mathrm{sec}$
4 $0.4 \mathrm{sec}$
Explanation:
C Given that, Magnetic flux, $\phi=3 \times 10^{-4} \mathrm{~Wb}$ $\mathrm{N}=100 \text { turns }$ EMF, $\varepsilon=1.5 \mathrm{~V}$ We know that, $|\varepsilon| =\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\mathrm{dt} =\frac{\mathrm{Nd} \phi}{\varepsilon}$ $\mathrm{dt} =\frac{100 \times 3 \times 10^{-4}}{1.5}$ $\mathrm{dt} =2 \times 10^{-2} \mathrm{~s}$ $\mathrm{dt} =0.02 \mathrm{sec}$ Time interval will be $0.02 \mathrm{sec}$.
COMEDK 2016]
Electro Magnetic Induction
154572
Assertion: An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf. Reason: Efficiency of electric motor depends only on magnitude of back emf.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is
Explanation:
C Efficiency $(\eta)=\frac{P_{\text {output }}}{P_{\text {input }}}$ Output power is given by $\varepsilon \mathrm{I}=\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}$ To obtain the maximum power differentiating $\mathrm{eq}^{\mathrm{n}}$ (i) with respect $\mathrm{e}$ and equating to zero $\frac{\mathrm{d}}{\mathrm{d} \varepsilon}\left[\frac{\varepsilon(\mathrm{E}-\varepsilon)}{\mathrm{R}}\right]=0$ $\frac{1}{\mathrm{R}}[\mathrm{E}-2 \varepsilon]=0$ $\mathrm{E}=2 \varepsilon$ $\varepsilon=\frac{\mathrm{E}}{2}$ Thus back emf $(\varepsilon)$ becomes equal to half of the applied voltage
AIIMS-2008
Electro Magnetic Induction
154573
The flux linked with a coil at any instant ' $t$ ' is given by $\phi=10 t^{2}-50 t+250$. The induced emf at $\mathbf{t}=\mathbf{3 s}$ is
1 $-190 \mathrm{~V}$
2 $-10 \mathrm{~V}$
3 $10 \mathrm{~V}$
4 $190 \mathrm{~V}$
Explanation:
B Given that, $\phi=10 \mathrm{t}^{2}-50 \mathrm{t}+250$ We know that, induced emf $\varepsilon= =\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon =\frac{-\mathrm{d}}{\mathrm{dt}}\left(10 \mathrm{t}^{2}-50 \mathrm{t}+250\right)$ $\varepsilon =-(20 \mathrm{t}-50)$ At, time $(\mathrm{t})=3 \mathrm{sec}$ then, $\varepsilon=-(20 \times 3-50)$ $\varepsilon=-60+50$ $\varepsilon=-10 \mathrm{~V}$
