154532
The magnetic flux linked with a closed coil is increased to a maximum value in $2 \mathrm{~s}$ and its relation with time is $\phi=a^{2}+b t+c$ then the relation between $a, b$ and $c$ is
154533
If $B_{H}=4 \times 10^{-5} \mathrm{~T}$ and $B_{v}=2 \times 10^{-5} \mathrm{~T}$, then the Earth's total field at the place is (in T)
1 $6 \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-5} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Explanation:
B Given that, $\mathrm{B}_{\mathrm{H}}=4 \times 10^{-5} \mathrm{~T}$ $\mathrm{B}_{\mathrm{V}}=2 \times 10^{-5} \mathrm{~T}$ We know that, The total earth's field at the place- $\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{V}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}$ $=\sqrt{\left(4 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-5}\right)^{2}}$ $=\sqrt{16 \times 10^{-10}+4 \times 10^{-10}}$ $=\sqrt{20} \times 10^{-5}=2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
J and K CET-2017
Electro Magnetic Induction
154534
A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is :
1 $400 \mathrm{mV}$
2 $200 \mathrm{mV}$
3 $4 \mathrm{mV}$
4 $0.4 \mathrm{mV}$
Explanation:
A Given that, No. of turns $(\mathrm{N})=1000$ Face area $(A)=4 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2}$ Change in magnetic field - $\Delta \mathrm{B}=10^{-2} \mathrm{Wbm}^{-2}$ Time taken, $\mathrm{t}=0.01 \mathrm{~s}=10^{-2} \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\mathrm{N}\left(\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\right) \mathrm{A}$ $\varepsilon=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}$ $\varepsilon=400 \mathrm{mV} .$
AIIMS-2017
Electro Magnetic Induction
154536
The magnetic flux through a coil varies with time as $\phi=5 t^{2}+6 t+9$. The ratio of emf at $t=$ $3 s$ to $t=0$ s will be
1 $1: 9$
2 $1: 6$
3 $6: 1$
4 $9: 1$
Explanation:
C We know that, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+6 \mathrm{t}+9\right)$ $\varepsilon=10 \mathrm{t}+6$ $\text { At } \mathrm{t}=3 \mathrm{sec}$ $\varepsilon_{1}=10 \times 3+6=36 \mathrm{~V}$ $\text { At } \mathrm{t}=0 \mathrm{sec}$ $\varepsilon_{2}=10 \times 0+6=6 \mathrm{~V}$ $\text { Therefore, } \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{36}{6}=\frac{6}{1}=6: 1 \mathrm{~V}$
154532
The magnetic flux linked with a closed coil is increased to a maximum value in $2 \mathrm{~s}$ and its relation with time is $\phi=a^{2}+b t+c$ then the relation between $a, b$ and $c$ is
154533
If $B_{H}=4 \times 10^{-5} \mathrm{~T}$ and $B_{v}=2 \times 10^{-5} \mathrm{~T}$, then the Earth's total field at the place is (in T)
1 $6 \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-5} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Explanation:
B Given that, $\mathrm{B}_{\mathrm{H}}=4 \times 10^{-5} \mathrm{~T}$ $\mathrm{B}_{\mathrm{V}}=2 \times 10^{-5} \mathrm{~T}$ We know that, The total earth's field at the place- $\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{V}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}$ $=\sqrt{\left(4 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-5}\right)^{2}}$ $=\sqrt{16 \times 10^{-10}+4 \times 10^{-10}}$ $=\sqrt{20} \times 10^{-5}=2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
J and K CET-2017
Electro Magnetic Induction
154534
A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is :
1 $400 \mathrm{mV}$
2 $200 \mathrm{mV}$
3 $4 \mathrm{mV}$
4 $0.4 \mathrm{mV}$
Explanation:
A Given that, No. of turns $(\mathrm{N})=1000$ Face area $(A)=4 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2}$ Change in magnetic field - $\Delta \mathrm{B}=10^{-2} \mathrm{Wbm}^{-2}$ Time taken, $\mathrm{t}=0.01 \mathrm{~s}=10^{-2} \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\mathrm{N}\left(\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\right) \mathrm{A}$ $\varepsilon=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}$ $\varepsilon=400 \mathrm{mV} .$
AIIMS-2017
Electro Magnetic Induction
154536
The magnetic flux through a coil varies with time as $\phi=5 t^{2}+6 t+9$. The ratio of emf at $t=$ $3 s$ to $t=0$ s will be
1 $1: 9$
2 $1: 6$
3 $6: 1$
4 $9: 1$
Explanation:
C We know that, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+6 \mathrm{t}+9\right)$ $\varepsilon=10 \mathrm{t}+6$ $\text { At } \mathrm{t}=3 \mathrm{sec}$ $\varepsilon_{1}=10 \times 3+6=36 \mathrm{~V}$ $\text { At } \mathrm{t}=0 \mathrm{sec}$ $\varepsilon_{2}=10 \times 0+6=6 \mathrm{~V}$ $\text { Therefore, } \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{36}{6}=\frac{6}{1}=6: 1 \mathrm{~V}$
154532
The magnetic flux linked with a closed coil is increased to a maximum value in $2 \mathrm{~s}$ and its relation with time is $\phi=a^{2}+b t+c$ then the relation between $a, b$ and $c$ is
154533
If $B_{H}=4 \times 10^{-5} \mathrm{~T}$ and $B_{v}=2 \times 10^{-5} \mathrm{~T}$, then the Earth's total field at the place is (in T)
1 $6 \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-5} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Explanation:
B Given that, $\mathrm{B}_{\mathrm{H}}=4 \times 10^{-5} \mathrm{~T}$ $\mathrm{B}_{\mathrm{V}}=2 \times 10^{-5} \mathrm{~T}$ We know that, The total earth's field at the place- $\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{V}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}$ $=\sqrt{\left(4 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-5}\right)^{2}}$ $=\sqrt{16 \times 10^{-10}+4 \times 10^{-10}}$ $=\sqrt{20} \times 10^{-5}=2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
J and K CET-2017
Electro Magnetic Induction
154534
A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is :
1 $400 \mathrm{mV}$
2 $200 \mathrm{mV}$
3 $4 \mathrm{mV}$
4 $0.4 \mathrm{mV}$
Explanation:
A Given that, No. of turns $(\mathrm{N})=1000$ Face area $(A)=4 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2}$ Change in magnetic field - $\Delta \mathrm{B}=10^{-2} \mathrm{Wbm}^{-2}$ Time taken, $\mathrm{t}=0.01 \mathrm{~s}=10^{-2} \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\mathrm{N}\left(\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\right) \mathrm{A}$ $\varepsilon=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}$ $\varepsilon=400 \mathrm{mV} .$
AIIMS-2017
Electro Magnetic Induction
154536
The magnetic flux through a coil varies with time as $\phi=5 t^{2}+6 t+9$. The ratio of emf at $t=$ $3 s$ to $t=0$ s will be
1 $1: 9$
2 $1: 6$
3 $6: 1$
4 $9: 1$
Explanation:
C We know that, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+6 \mathrm{t}+9\right)$ $\varepsilon=10 \mathrm{t}+6$ $\text { At } \mathrm{t}=3 \mathrm{sec}$ $\varepsilon_{1}=10 \times 3+6=36 \mathrm{~V}$ $\text { At } \mathrm{t}=0 \mathrm{sec}$ $\varepsilon_{2}=10 \times 0+6=6 \mathrm{~V}$ $\text { Therefore, } \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{36}{6}=\frac{6}{1}=6: 1 \mathrm{~V}$
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Electro Magnetic Induction
154532
The magnetic flux linked with a closed coil is increased to a maximum value in $2 \mathrm{~s}$ and its relation with time is $\phi=a^{2}+b t+c$ then the relation between $a, b$ and $c$ is
154533
If $B_{H}=4 \times 10^{-5} \mathrm{~T}$ and $B_{v}=2 \times 10^{-5} \mathrm{~T}$, then the Earth's total field at the place is (in T)
1 $6 \times 10^{-5} \mathrm{~T}$
2 $2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
3 $4 \times 10^{-5} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Explanation:
B Given that, $\mathrm{B}_{\mathrm{H}}=4 \times 10^{-5} \mathrm{~T}$ $\mathrm{B}_{\mathrm{V}}=2 \times 10^{-5} \mathrm{~T}$ We know that, The total earth's field at the place- $\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{V}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}$ $=\sqrt{\left(4 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-5}\right)^{2}}$ $=\sqrt{16 \times 10^{-10}+4 \times 10^{-10}}$ $=\sqrt{20} \times 10^{-5}=2 \sqrt{5} \times 10^{-5} \mathrm{~T}$
J and K CET-2017
Electro Magnetic Induction
154534
A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is :
1 $400 \mathrm{mV}$
2 $200 \mathrm{mV}$
3 $4 \mathrm{mV}$
4 $0.4 \mathrm{mV}$
Explanation:
A Given that, No. of turns $(\mathrm{N})=1000$ Face area $(A)=4 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2}$ Change in magnetic field - $\Delta \mathrm{B}=10^{-2} \mathrm{Wbm}^{-2}$ Time taken, $\mathrm{t}=0.01 \mathrm{~s}=10^{-2} \mathrm{~s}$ We know that, Induced emf $(\varepsilon)=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\mathrm{N}\left(\frac{\Delta \mathrm{B}}{\Delta \mathrm{t}}\right) \mathrm{A}$ $\varepsilon=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}$ $\varepsilon=400 \mathrm{mV} .$
AIIMS-2017
Electro Magnetic Induction
154536
The magnetic flux through a coil varies with time as $\phi=5 t^{2}+6 t+9$. The ratio of emf at $t=$ $3 s$ to $t=0$ s will be
1 $1: 9$
2 $1: 6$
3 $6: 1$
4 $9: 1$
Explanation:
C We know that, $|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}$ $\varepsilon=\frac{\mathrm{d}}{\mathrm{dt}}\left(5 \mathrm{t}^{2}+6 \mathrm{t}+9\right)$ $\varepsilon=10 \mathrm{t}+6$ $\text { At } \mathrm{t}=3 \mathrm{sec}$ $\varepsilon_{1}=10 \times 3+6=36 \mathrm{~V}$ $\text { At } \mathrm{t}=0 \mathrm{sec}$ $\varepsilon_{2}=10 \times 0+6=6 \mathrm{~V}$ $\text { Therefore, } \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{36}{6}=\frac{6}{1}=6: 1 \mathrm{~V}$
