NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154870
A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \times$ $0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. The induced emf when the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$ is-
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, Number of turns $\mathrm{N}=100$ turns Area $\mathrm{A}=0.1 \times 0.05 \mathrm{~m}$ Primary magnetic field $\mathrm{B}_{1}=0.1 \mathrm{~T}$ Secondary magnetic field $\mathrm{B}_{2}=0.05 \mathrm{~T}$ Time $\Delta \mathrm{T}=0.05 \mathrm{sec}$ $\because \operatorname{emf} \varepsilon=\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\therefore$ emf $\varepsilon=\mathrm{N} \frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=100 \times 0.1 \times 0.05 \times \frac{(0.1-0.05)}{0.05}$ $\operatorname{emf} \varepsilon=0.5 \mathrm{~V}$
BCECE-2009
Electro Magnetic Induction
154871
A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends of the rod is-
1 $18.89 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $9.42 \mathrm{~V}$
Explanation:
D Given that, Length of $\operatorname{rod} l=1.0 \mathrm{~m}$ Magnetic field, $\mathrm{B}=0.25 \mathrm{~T}$ Frequency, $\omega=12 \mathrm{rev} / \mathrm{sec}$ $\operatorname{emf} \varepsilon=\mathrm{BA} \omega$ emf $\varepsilon=0.25 \times \pi(1)^{2} \times 12$ emf $\varepsilon=9.42 \mathrm{~V}$
AP EAMCET (22.09.2020) Shift-II
Electro Magnetic Induction
154872
The inductance of a coil is $L=10 \mathrm{H}$ and resistance $R=5 \Omega$. If applied voltage of battery is $10 \mathrm{~V}$ and it switches off in 1 millisecond, find induced emf of inductor.
1 $2 \times 10^{4} \mathrm{~V}$
2 $1.2 \times 10^{4} \mathrm{~V}$
3 $2 \times 10^{-4} \mathrm{~V}$
4 None of these
Explanation:
A Given that, Self inductance $\mathrm{L}=10 \mathrm{H}$, Resistance $\mathrm{R}=5 \Omega$ Voltage of battery $\mathrm{V}=10 \mathrm{~V}$ Time $\Delta \mathrm{t}=1 \times 10^{-3} \mathrm{sec}$ Induced current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{5}=2 \mathrm{~A}$ $\because \quad$ emf $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=10 \times \frac{2}{10^{-3}}$ $|\varepsilon|=2 \times 10^{4} \mathrm{~V}$
BCECE-2008
Electro Magnetic Induction
154873
The current in a coil decreases from $1 \mathrm{~A}$ to $0.2 \mathrm{~A}$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 \mathrm{~V}$, is :
1 $5 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
A Given that, Primary current $\mathrm{I}_{1}=1 \mathrm{~A}$, Secondary current $\mathrm{I}_{2}=0.2 \mathrm{~A}$ Time $\Delta \mathrm{t}=10 \mathrm{sec}$, emf $\varepsilon=0.4 \mathrm{~V}$ $\mathrm{L} =\frac{\mathrm{emf}}{\Delta \mathrm{I} / \Delta \mathrm{t}}$ $\mathrm{L} =\frac{0.4}{(1-0.2) / 10}=\frac{4}{0.8}=5 \mathrm{H}$ $\mathrm{L} =5 \mathrm{H}$
154870
A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \times$ $0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. The induced emf when the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$ is-
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, Number of turns $\mathrm{N}=100$ turns Area $\mathrm{A}=0.1 \times 0.05 \mathrm{~m}$ Primary magnetic field $\mathrm{B}_{1}=0.1 \mathrm{~T}$ Secondary magnetic field $\mathrm{B}_{2}=0.05 \mathrm{~T}$ Time $\Delta \mathrm{T}=0.05 \mathrm{sec}$ $\because \operatorname{emf} \varepsilon=\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\therefore$ emf $\varepsilon=\mathrm{N} \frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=100 \times 0.1 \times 0.05 \times \frac{(0.1-0.05)}{0.05}$ $\operatorname{emf} \varepsilon=0.5 \mathrm{~V}$
BCECE-2009
Electro Magnetic Induction
154871
A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends of the rod is-
1 $18.89 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $9.42 \mathrm{~V}$
Explanation:
D Given that, Length of $\operatorname{rod} l=1.0 \mathrm{~m}$ Magnetic field, $\mathrm{B}=0.25 \mathrm{~T}$ Frequency, $\omega=12 \mathrm{rev} / \mathrm{sec}$ $\operatorname{emf} \varepsilon=\mathrm{BA} \omega$ emf $\varepsilon=0.25 \times \pi(1)^{2} \times 12$ emf $\varepsilon=9.42 \mathrm{~V}$
AP EAMCET (22.09.2020) Shift-II
Electro Magnetic Induction
154872
The inductance of a coil is $L=10 \mathrm{H}$ and resistance $R=5 \Omega$. If applied voltage of battery is $10 \mathrm{~V}$ and it switches off in 1 millisecond, find induced emf of inductor.
1 $2 \times 10^{4} \mathrm{~V}$
2 $1.2 \times 10^{4} \mathrm{~V}$
3 $2 \times 10^{-4} \mathrm{~V}$
4 None of these
Explanation:
A Given that, Self inductance $\mathrm{L}=10 \mathrm{H}$, Resistance $\mathrm{R}=5 \Omega$ Voltage of battery $\mathrm{V}=10 \mathrm{~V}$ Time $\Delta \mathrm{t}=1 \times 10^{-3} \mathrm{sec}$ Induced current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{5}=2 \mathrm{~A}$ $\because \quad$ emf $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=10 \times \frac{2}{10^{-3}}$ $|\varepsilon|=2 \times 10^{4} \mathrm{~V}$
BCECE-2008
Electro Magnetic Induction
154873
The current in a coil decreases from $1 \mathrm{~A}$ to $0.2 \mathrm{~A}$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 \mathrm{~V}$, is :
1 $5 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
A Given that, Primary current $\mathrm{I}_{1}=1 \mathrm{~A}$, Secondary current $\mathrm{I}_{2}=0.2 \mathrm{~A}$ Time $\Delta \mathrm{t}=10 \mathrm{sec}$, emf $\varepsilon=0.4 \mathrm{~V}$ $\mathrm{L} =\frac{\mathrm{emf}}{\Delta \mathrm{I} / \Delta \mathrm{t}}$ $\mathrm{L} =\frac{0.4}{(1-0.2) / 10}=\frac{4}{0.8}=5 \mathrm{H}$ $\mathrm{L} =5 \mathrm{H}$
154870
A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \times$ $0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. The induced emf when the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$ is-
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, Number of turns $\mathrm{N}=100$ turns Area $\mathrm{A}=0.1 \times 0.05 \mathrm{~m}$ Primary magnetic field $\mathrm{B}_{1}=0.1 \mathrm{~T}$ Secondary magnetic field $\mathrm{B}_{2}=0.05 \mathrm{~T}$ Time $\Delta \mathrm{T}=0.05 \mathrm{sec}$ $\because \operatorname{emf} \varepsilon=\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\therefore$ emf $\varepsilon=\mathrm{N} \frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=100 \times 0.1 \times 0.05 \times \frac{(0.1-0.05)}{0.05}$ $\operatorname{emf} \varepsilon=0.5 \mathrm{~V}$
BCECE-2009
Electro Magnetic Induction
154871
A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends of the rod is-
1 $18.89 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $9.42 \mathrm{~V}$
Explanation:
D Given that, Length of $\operatorname{rod} l=1.0 \mathrm{~m}$ Magnetic field, $\mathrm{B}=0.25 \mathrm{~T}$ Frequency, $\omega=12 \mathrm{rev} / \mathrm{sec}$ $\operatorname{emf} \varepsilon=\mathrm{BA} \omega$ emf $\varepsilon=0.25 \times \pi(1)^{2} \times 12$ emf $\varepsilon=9.42 \mathrm{~V}$
AP EAMCET (22.09.2020) Shift-II
Electro Magnetic Induction
154872
The inductance of a coil is $L=10 \mathrm{H}$ and resistance $R=5 \Omega$. If applied voltage of battery is $10 \mathrm{~V}$ and it switches off in 1 millisecond, find induced emf of inductor.
1 $2 \times 10^{4} \mathrm{~V}$
2 $1.2 \times 10^{4} \mathrm{~V}$
3 $2 \times 10^{-4} \mathrm{~V}$
4 None of these
Explanation:
A Given that, Self inductance $\mathrm{L}=10 \mathrm{H}$, Resistance $\mathrm{R}=5 \Omega$ Voltage of battery $\mathrm{V}=10 \mathrm{~V}$ Time $\Delta \mathrm{t}=1 \times 10^{-3} \mathrm{sec}$ Induced current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{5}=2 \mathrm{~A}$ $\because \quad$ emf $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=10 \times \frac{2}{10^{-3}}$ $|\varepsilon|=2 \times 10^{4} \mathrm{~V}$
BCECE-2008
Electro Magnetic Induction
154873
The current in a coil decreases from $1 \mathrm{~A}$ to $0.2 \mathrm{~A}$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 \mathrm{~V}$, is :
1 $5 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
A Given that, Primary current $\mathrm{I}_{1}=1 \mathrm{~A}$, Secondary current $\mathrm{I}_{2}=0.2 \mathrm{~A}$ Time $\Delta \mathrm{t}=10 \mathrm{sec}$, emf $\varepsilon=0.4 \mathrm{~V}$ $\mathrm{L} =\frac{\mathrm{emf}}{\Delta \mathrm{I} / \Delta \mathrm{t}}$ $\mathrm{L} =\frac{0.4}{(1-0.2) / 10}=\frac{4}{0.8}=5 \mathrm{H}$ $\mathrm{L} =5 \mathrm{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electro Magnetic Induction
154870
A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \times$ $0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. The induced emf when the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$ is-
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, Number of turns $\mathrm{N}=100$ turns Area $\mathrm{A}=0.1 \times 0.05 \mathrm{~m}$ Primary magnetic field $\mathrm{B}_{1}=0.1 \mathrm{~T}$ Secondary magnetic field $\mathrm{B}_{2}=0.05 \mathrm{~T}$ Time $\Delta \mathrm{T}=0.05 \mathrm{sec}$ $\because \operatorname{emf} \varepsilon=\frac{\mathrm{Nd} \phi}{\mathrm{dt}}$ $\therefore$ emf $\varepsilon=\mathrm{N} \frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=100 \times 0.1 \times 0.05 \times \frac{(0.1-0.05)}{0.05}$ $\operatorname{emf} \varepsilon=0.5 \mathrm{~V}$
BCECE-2009
Electro Magnetic Induction
154871
A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends of the rod is-
1 $18.89 \mathrm{~V}$
2 $3 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 $9.42 \mathrm{~V}$
Explanation:
D Given that, Length of $\operatorname{rod} l=1.0 \mathrm{~m}$ Magnetic field, $\mathrm{B}=0.25 \mathrm{~T}$ Frequency, $\omega=12 \mathrm{rev} / \mathrm{sec}$ $\operatorname{emf} \varepsilon=\mathrm{BA} \omega$ emf $\varepsilon=0.25 \times \pi(1)^{2} \times 12$ emf $\varepsilon=9.42 \mathrm{~V}$
AP EAMCET (22.09.2020) Shift-II
Electro Magnetic Induction
154872
The inductance of a coil is $L=10 \mathrm{H}$ and resistance $R=5 \Omega$. If applied voltage of battery is $10 \mathrm{~V}$ and it switches off in 1 millisecond, find induced emf of inductor.
1 $2 \times 10^{4} \mathrm{~V}$
2 $1.2 \times 10^{4} \mathrm{~V}$
3 $2 \times 10^{-4} \mathrm{~V}$
4 None of these
Explanation:
A Given that, Self inductance $\mathrm{L}=10 \mathrm{H}$, Resistance $\mathrm{R}=5 \Omega$ Voltage of battery $\mathrm{V}=10 \mathrm{~V}$ Time $\Delta \mathrm{t}=1 \times 10^{-3} \mathrm{sec}$ Induced current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{5}=2 \mathrm{~A}$ $\because \quad$ emf $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\therefore \quad$ emf $\varepsilon=10 \times \frac{2}{10^{-3}}$ $|\varepsilon|=2 \times 10^{4} \mathrm{~V}$
BCECE-2008
Electro Magnetic Induction
154873
The current in a coil decreases from $1 \mathrm{~A}$ to $0.2 \mathrm{~A}$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 \mathrm{~V}$, is :
1 $5 \mathrm{H}$
2 $3 \mathrm{H}$
3 $4 \mathrm{H}$
4 $2 \mathrm{H}$
Explanation:
A Given that, Primary current $\mathrm{I}_{1}=1 \mathrm{~A}$, Secondary current $\mathrm{I}_{2}=0.2 \mathrm{~A}$ Time $\Delta \mathrm{t}=10 \mathrm{sec}$, emf $\varepsilon=0.4 \mathrm{~V}$ $\mathrm{L} =\frac{\mathrm{emf}}{\Delta \mathrm{I} / \Delta \mathrm{t}}$ $\mathrm{L} =\frac{0.4}{(1-0.2) / 10}=\frac{4}{0.8}=5 \mathrm{H}$ $\mathrm{L} =5 \mathrm{H}$