154751
A rod of length $1 \mathrm{~m}$ is kept inclined at an angle of $60^{\circ}$ with the uniform magnetic field of $0.5 \mathrm{~T}$. If the rod is moved with a velocity $10 \mathrm{~m} . \mathrm{s}^{-1}$ perpendicular to the field, the induced emf is:
1 $10 \mathrm{~V}$
2 $7.5 \mathrm{~V}$
3 $4.33 \mathrm{~V}$
4 $2.55 \mathrm{~V}$
Explanation:
C Given, Length of $\operatorname{Rod}(l)=1 \mathrm{~m}$ Magnetic field $(\mathrm{B})=0.5 \mathrm{~T}$ Velocity $(\mathrm{v})=10 \mathrm{~m} / \mathrm{s}$ Induced emf $(\varepsilon)=\mathrm{B} l \mathrm{v} \sin \theta$ $\varepsilon=0.5 \times 1 \times 10 \times \sin 60^{\circ}$ $\varepsilon=5 \times \frac{\sqrt{3}}{2}=4.33 \mathrm{~V}$
AP EAMCET-23.09.2020
Electro Magnetic Induction
154752
When the current in a coil changes from $4 \mathrm{~A}$ to $8 \mathrm{~A}$ in $0.1 \mathrm{~s}$, an e.m.f. of $16 \mathrm{~V}$ is induced in it. Then find the coefficient of self-induction of the coil.
1 $0.2 \mathrm{H}$
2 $0.35 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.48 \mathrm{H}$
Explanation:
C Given, Change in current $=8-4=4 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=16 \mathrm{~V}$ $\therefore \quad \varepsilon =\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L} =\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{16 \times 0.1}{4}$ $\mathrm{~L} =4 \times 0.1=0.4 \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154753
The back e.m.f. induced in a coil, when current changes from $1 \mathrm{~A}$ to $0 \mathrm{~A}$ in 1 milli-second is 4 $\mathrm{V}$. The self-inductance of the coil is
1 $1 \mathrm{H}$
2 $4 \mathrm{H}$
3 $10^{-3} \mathrm{H}$
4 $4 \times 10^{-3} \mathrm{H}$
Explanation:
D Emf induced $(\varepsilon)=4 \mathrm{~V}$ Change in current $(\Delta \mathrm{I})=(1-0)=1 \mathrm{~A}$ Time $(\Delta \mathrm{t})=1 \mathrm{~m}-\mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}$ $\therefore \quad \mathrm{L}=4 \times \frac{10^{-3}}{1}=4 \times 10^{-3} \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154754
The average emf induced in a coil, when the current in it changes from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in $0.05 \mathrm{~s}$. is 8V. Find the self-inductance of this coil.
1 $0.8 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.2 \mathrm{H}$
4 $0.1 \mathrm{H}$
Explanation:
C Given, Induced emf $(\varepsilon)=8 \mathrm{~V}$ Change in current $=4-2=2 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.05 \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\therefore \quad \mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=8 \times \frac{0.05}{2}$ $\mathrm{L}=4 \times 0.05=0.2 \mathrm{H}$
154751
A rod of length $1 \mathrm{~m}$ is kept inclined at an angle of $60^{\circ}$ with the uniform magnetic field of $0.5 \mathrm{~T}$. If the rod is moved with a velocity $10 \mathrm{~m} . \mathrm{s}^{-1}$ perpendicular to the field, the induced emf is:
1 $10 \mathrm{~V}$
2 $7.5 \mathrm{~V}$
3 $4.33 \mathrm{~V}$
4 $2.55 \mathrm{~V}$
Explanation:
C Given, Length of $\operatorname{Rod}(l)=1 \mathrm{~m}$ Magnetic field $(\mathrm{B})=0.5 \mathrm{~T}$ Velocity $(\mathrm{v})=10 \mathrm{~m} / \mathrm{s}$ Induced emf $(\varepsilon)=\mathrm{B} l \mathrm{v} \sin \theta$ $\varepsilon=0.5 \times 1 \times 10 \times \sin 60^{\circ}$ $\varepsilon=5 \times \frac{\sqrt{3}}{2}=4.33 \mathrm{~V}$
AP EAMCET-23.09.2020
Electro Magnetic Induction
154752
When the current in a coil changes from $4 \mathrm{~A}$ to $8 \mathrm{~A}$ in $0.1 \mathrm{~s}$, an e.m.f. of $16 \mathrm{~V}$ is induced in it. Then find the coefficient of self-induction of the coil.
1 $0.2 \mathrm{H}$
2 $0.35 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.48 \mathrm{H}$
Explanation:
C Given, Change in current $=8-4=4 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=16 \mathrm{~V}$ $\therefore \quad \varepsilon =\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L} =\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{16 \times 0.1}{4}$ $\mathrm{~L} =4 \times 0.1=0.4 \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154753
The back e.m.f. induced in a coil, when current changes from $1 \mathrm{~A}$ to $0 \mathrm{~A}$ in 1 milli-second is 4 $\mathrm{V}$. The self-inductance of the coil is
1 $1 \mathrm{H}$
2 $4 \mathrm{H}$
3 $10^{-3} \mathrm{H}$
4 $4 \times 10^{-3} \mathrm{H}$
Explanation:
D Emf induced $(\varepsilon)=4 \mathrm{~V}$ Change in current $(\Delta \mathrm{I})=(1-0)=1 \mathrm{~A}$ Time $(\Delta \mathrm{t})=1 \mathrm{~m}-\mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}$ $\therefore \quad \mathrm{L}=4 \times \frac{10^{-3}}{1}=4 \times 10^{-3} \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154754
The average emf induced in a coil, when the current in it changes from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in $0.05 \mathrm{~s}$. is 8V. Find the self-inductance of this coil.
1 $0.8 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.2 \mathrm{H}$
4 $0.1 \mathrm{H}$
Explanation:
C Given, Induced emf $(\varepsilon)=8 \mathrm{~V}$ Change in current $=4-2=2 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.05 \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\therefore \quad \mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=8 \times \frac{0.05}{2}$ $\mathrm{L}=4 \times 0.05=0.2 \mathrm{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154751
A rod of length $1 \mathrm{~m}$ is kept inclined at an angle of $60^{\circ}$ with the uniform magnetic field of $0.5 \mathrm{~T}$. If the rod is moved with a velocity $10 \mathrm{~m} . \mathrm{s}^{-1}$ perpendicular to the field, the induced emf is:
1 $10 \mathrm{~V}$
2 $7.5 \mathrm{~V}$
3 $4.33 \mathrm{~V}$
4 $2.55 \mathrm{~V}$
Explanation:
C Given, Length of $\operatorname{Rod}(l)=1 \mathrm{~m}$ Magnetic field $(\mathrm{B})=0.5 \mathrm{~T}$ Velocity $(\mathrm{v})=10 \mathrm{~m} / \mathrm{s}$ Induced emf $(\varepsilon)=\mathrm{B} l \mathrm{v} \sin \theta$ $\varepsilon=0.5 \times 1 \times 10 \times \sin 60^{\circ}$ $\varepsilon=5 \times \frac{\sqrt{3}}{2}=4.33 \mathrm{~V}$
AP EAMCET-23.09.2020
Electro Magnetic Induction
154752
When the current in a coil changes from $4 \mathrm{~A}$ to $8 \mathrm{~A}$ in $0.1 \mathrm{~s}$, an e.m.f. of $16 \mathrm{~V}$ is induced in it. Then find the coefficient of self-induction of the coil.
1 $0.2 \mathrm{H}$
2 $0.35 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.48 \mathrm{H}$
Explanation:
C Given, Change in current $=8-4=4 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=16 \mathrm{~V}$ $\therefore \quad \varepsilon =\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L} =\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{16 \times 0.1}{4}$ $\mathrm{~L} =4 \times 0.1=0.4 \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154753
The back e.m.f. induced in a coil, when current changes from $1 \mathrm{~A}$ to $0 \mathrm{~A}$ in 1 milli-second is 4 $\mathrm{V}$. The self-inductance of the coil is
1 $1 \mathrm{H}$
2 $4 \mathrm{H}$
3 $10^{-3} \mathrm{H}$
4 $4 \times 10^{-3} \mathrm{H}$
Explanation:
D Emf induced $(\varepsilon)=4 \mathrm{~V}$ Change in current $(\Delta \mathrm{I})=(1-0)=1 \mathrm{~A}$ Time $(\Delta \mathrm{t})=1 \mathrm{~m}-\mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}$ $\therefore \quad \mathrm{L}=4 \times \frac{10^{-3}}{1}=4 \times 10^{-3} \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154754
The average emf induced in a coil, when the current in it changes from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in $0.05 \mathrm{~s}$. is 8V. Find the self-inductance of this coil.
1 $0.8 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.2 \mathrm{H}$
4 $0.1 \mathrm{H}$
Explanation:
C Given, Induced emf $(\varepsilon)=8 \mathrm{~V}$ Change in current $=4-2=2 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.05 \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\therefore \quad \mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=8 \times \frac{0.05}{2}$ $\mathrm{L}=4 \times 0.05=0.2 \mathrm{H}$
154751
A rod of length $1 \mathrm{~m}$ is kept inclined at an angle of $60^{\circ}$ with the uniform magnetic field of $0.5 \mathrm{~T}$. If the rod is moved with a velocity $10 \mathrm{~m} . \mathrm{s}^{-1}$ perpendicular to the field, the induced emf is:
1 $10 \mathrm{~V}$
2 $7.5 \mathrm{~V}$
3 $4.33 \mathrm{~V}$
4 $2.55 \mathrm{~V}$
Explanation:
C Given, Length of $\operatorname{Rod}(l)=1 \mathrm{~m}$ Magnetic field $(\mathrm{B})=0.5 \mathrm{~T}$ Velocity $(\mathrm{v})=10 \mathrm{~m} / \mathrm{s}$ Induced emf $(\varepsilon)=\mathrm{B} l \mathrm{v} \sin \theta$ $\varepsilon=0.5 \times 1 \times 10 \times \sin 60^{\circ}$ $\varepsilon=5 \times \frac{\sqrt{3}}{2}=4.33 \mathrm{~V}$
AP EAMCET-23.09.2020
Electro Magnetic Induction
154752
When the current in a coil changes from $4 \mathrm{~A}$ to $8 \mathrm{~A}$ in $0.1 \mathrm{~s}$, an e.m.f. of $16 \mathrm{~V}$ is induced in it. Then find the coefficient of self-induction of the coil.
1 $0.2 \mathrm{H}$
2 $0.35 \mathrm{H}$
3 $0.4 \mathrm{H}$
4 $0.48 \mathrm{H}$
Explanation:
C Given, Change in current $=8-4=4 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.1 \mathrm{sec}$ emf induced $(\varepsilon)=16 \mathrm{~V}$ $\therefore \quad \varepsilon =\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L} =\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=\frac{16 \times 0.1}{4}$ $\mathrm{~L} =4 \times 0.1=0.4 \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154753
The back e.m.f. induced in a coil, when current changes from $1 \mathrm{~A}$ to $0 \mathrm{~A}$ in 1 milli-second is 4 $\mathrm{V}$. The self-inductance of the coil is
1 $1 \mathrm{H}$
2 $4 \mathrm{H}$
3 $10^{-3} \mathrm{H}$
4 $4 \times 10^{-3} \mathrm{H}$
Explanation:
D Emf induced $(\varepsilon)=4 \mathrm{~V}$ Change in current $(\Delta \mathrm{I})=(1-0)=1 \mathrm{~A}$ Time $(\Delta \mathrm{t})=1 \mathrm{~m}-\mathrm{sec}=1 \times 10^{-3} \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}$ $\therefore \quad \mathrm{L}=4 \times \frac{10^{-3}}{1}=4 \times 10^{-3} \mathrm{H}$
AP EAMCET-25.09.2020
Electro Magnetic Induction
154754
The average emf induced in a coil, when the current in it changes from $2 \mathrm{~A}$ to $4 \mathrm{~A}$ in $0.05 \mathrm{~s}$. is 8V. Find the self-inductance of this coil.
1 $0.8 \mathrm{H}$
2 $0.4 \mathrm{H}$
3 $0.2 \mathrm{H}$
4 $0.1 \mathrm{H}$
Explanation:
C Given, Induced emf $(\varepsilon)=8 \mathrm{~V}$ Change in current $=4-2=2 \mathrm{~A}$ Time $(\Delta \mathrm{t})=0.05 \mathrm{sec}$ Induced emf $(\varepsilon)=\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}$ $\therefore \quad \mathrm{L}=\varepsilon \frac{\Delta \mathrm{t}}{\Delta \mathrm{I}}=8 \times \frac{0.05}{2}$ $\mathrm{L}=4 \times 0.05=0.2 \mathrm{H}$
