NEET Test Series from KOTA - 10 Papers In MS WORD
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Electro Magnetic Induction
154784
Consider a current in a circuit falls from 6.0 $\mathrm{A}$ to $1.0 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If an average emf of $150 \mathrm{~V}$ is induced by the circuit, then the self inductance of the circuit is
154785
Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying n number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is
A Given, Inner radius $=\mathrm{a}$ Outer radius $=\mathrm{b}$ Height $=\mathrm{h}$ We know that, magnetic field inside a rectangular toroid is- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$ Let, $\mathrm{dx}$ is a small element of cross-section $\therefore \quad \mathrm{dx}=\mathrm{hdr}$ So, the magnetic flux through the cross-section of toroid $\phi=\int \mathrm{B} \cdot \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$. (h.dr) $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi} \int_{\mathrm{a}}^{\mathrm{b}} \frac{1}{\mathrm{r}} \mathrm{dr} .$ $\phi=\frac{\mu_{0} \mathrm{nIh}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln \mathrm{b}-\ln \mathrm{a}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln (\mathrm{b} / \mathrm{a})]$ $\because$ Self inductance $(L)=\frac{\mathrm{n} \phi}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_{\mathrm{o}} \mathrm{n}^{2} \mathrm{~h}}{2 \pi} \ln (\mathrm{b} / \mathrm{a})$
TS- EAMCET-06.05.2019
Electro Magnetic Induction
154786
A coil is connected to an AC source with peak emf, $8 V$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The coil has resistance of $8 \Omega$. If the average power dissipated by the coil is $0.4 \mathrm{~W}$, then the inductance of the coil is
154787
The magnetic potential energy stored in a certain inductor is $25 \mathrm{~mJ}$, when the current in the inductor is $60 \mathrm{~mA}$. This inductor is of inductance
1 $1.389 \mathrm{H}$
2 $138.88 \mathrm{H}$
3 $0.138 \mathrm{H}$
4 $13.89 \mathrm{H}$
Explanation:
D Given, $\mathrm{U}=25 \mathrm{~mJ}=25 \times 10^{-3} \mathrm{~J}, \mathrm{I}=60 \mathrm{~mA}=$ $60 \times 10^{-3} \mathrm{~A}$ We know that, energy stored in inductor is, $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{~L}=\frac{2 \mathrm{U}}{\mathrm{I}^{2}}$ $\mathrm{~L}=\frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^{2}}$ $\mathrm{~L}=13.89 \mathrm{H}$
154784
Consider a current in a circuit falls from 6.0 $\mathrm{A}$ to $1.0 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If an average emf of $150 \mathrm{~V}$ is induced by the circuit, then the self inductance of the circuit is
154785
Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying n number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is
A Given, Inner radius $=\mathrm{a}$ Outer radius $=\mathrm{b}$ Height $=\mathrm{h}$ We know that, magnetic field inside a rectangular toroid is- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$ Let, $\mathrm{dx}$ is a small element of cross-section $\therefore \quad \mathrm{dx}=\mathrm{hdr}$ So, the magnetic flux through the cross-section of toroid $\phi=\int \mathrm{B} \cdot \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$. (h.dr) $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi} \int_{\mathrm{a}}^{\mathrm{b}} \frac{1}{\mathrm{r}} \mathrm{dr} .$ $\phi=\frac{\mu_{0} \mathrm{nIh}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln \mathrm{b}-\ln \mathrm{a}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln (\mathrm{b} / \mathrm{a})]$ $\because$ Self inductance $(L)=\frac{\mathrm{n} \phi}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_{\mathrm{o}} \mathrm{n}^{2} \mathrm{~h}}{2 \pi} \ln (\mathrm{b} / \mathrm{a})$
TS- EAMCET-06.05.2019
Electro Magnetic Induction
154786
A coil is connected to an AC source with peak emf, $8 V$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The coil has resistance of $8 \Omega$. If the average power dissipated by the coil is $0.4 \mathrm{~W}$, then the inductance of the coil is
154787
The magnetic potential energy stored in a certain inductor is $25 \mathrm{~mJ}$, when the current in the inductor is $60 \mathrm{~mA}$. This inductor is of inductance
1 $1.389 \mathrm{H}$
2 $138.88 \mathrm{H}$
3 $0.138 \mathrm{H}$
4 $13.89 \mathrm{H}$
Explanation:
D Given, $\mathrm{U}=25 \mathrm{~mJ}=25 \times 10^{-3} \mathrm{~J}, \mathrm{I}=60 \mathrm{~mA}=$ $60 \times 10^{-3} \mathrm{~A}$ We know that, energy stored in inductor is, $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{~L}=\frac{2 \mathrm{U}}{\mathrm{I}^{2}}$ $\mathrm{~L}=\frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^{2}}$ $\mathrm{~L}=13.89 \mathrm{H}$
154784
Consider a current in a circuit falls from 6.0 $\mathrm{A}$ to $1.0 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If an average emf of $150 \mathrm{~V}$ is induced by the circuit, then the self inductance of the circuit is
154785
Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying n number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is
A Given, Inner radius $=\mathrm{a}$ Outer radius $=\mathrm{b}$ Height $=\mathrm{h}$ We know that, magnetic field inside a rectangular toroid is- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$ Let, $\mathrm{dx}$ is a small element of cross-section $\therefore \quad \mathrm{dx}=\mathrm{hdr}$ So, the magnetic flux through the cross-section of toroid $\phi=\int \mathrm{B} \cdot \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$. (h.dr) $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi} \int_{\mathrm{a}}^{\mathrm{b}} \frac{1}{\mathrm{r}} \mathrm{dr} .$ $\phi=\frac{\mu_{0} \mathrm{nIh}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln \mathrm{b}-\ln \mathrm{a}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln (\mathrm{b} / \mathrm{a})]$ $\because$ Self inductance $(L)=\frac{\mathrm{n} \phi}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_{\mathrm{o}} \mathrm{n}^{2} \mathrm{~h}}{2 \pi} \ln (\mathrm{b} / \mathrm{a})$
TS- EAMCET-06.05.2019
Electro Magnetic Induction
154786
A coil is connected to an AC source with peak emf, $8 V$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The coil has resistance of $8 \Omega$. If the average power dissipated by the coil is $0.4 \mathrm{~W}$, then the inductance of the coil is
154787
The magnetic potential energy stored in a certain inductor is $25 \mathrm{~mJ}$, when the current in the inductor is $60 \mathrm{~mA}$. This inductor is of inductance
1 $1.389 \mathrm{H}$
2 $138.88 \mathrm{H}$
3 $0.138 \mathrm{H}$
4 $13.89 \mathrm{H}$
Explanation:
D Given, $\mathrm{U}=25 \mathrm{~mJ}=25 \times 10^{-3} \mathrm{~J}, \mathrm{I}=60 \mathrm{~mA}=$ $60 \times 10^{-3} \mathrm{~A}$ We know that, energy stored in inductor is, $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{~L}=\frac{2 \mathrm{U}}{\mathrm{I}^{2}}$ $\mathrm{~L}=\frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^{2}}$ $\mathrm{~L}=13.89 \mathrm{H}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electro Magnetic Induction
154784
Consider a current in a circuit falls from 6.0 $\mathrm{A}$ to $1.0 \mathrm{~A}$ in $0.2 \mathrm{~s}$. If an average emf of $150 \mathrm{~V}$ is induced by the circuit, then the self inductance of the circuit is
154785
Consider a toroid with rectangular cross section, of inner radius $a$, outer radius $b$ and height $h$, carrying n number of turns. Then the self-inductance of the toroidal coil when current $I$ passing through the toroid is
A Given, Inner radius $=\mathrm{a}$ Outer radius $=\mathrm{b}$ Height $=\mathrm{h}$ We know that, magnetic field inside a rectangular toroid is- $\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$ Let, $\mathrm{dx}$ is a small element of cross-section $\therefore \quad \mathrm{dx}=\mathrm{hdr}$ So, the magnetic flux through the cross-section of toroid $\phi=\int \mathrm{B} \cdot \mathrm{dx}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{\mathrm{o}} \mathrm{nI}}{2 \pi \mathrm{r}}$. (h.dr) $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi} \int_{\mathrm{a}}^{\mathrm{b}} \frac{1}{\mathrm{r}} \mathrm{dr} .$ $\phi=\frac{\mu_{0} \mathrm{nIh}}{2 \pi}[\ln \mathrm{r}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln \mathrm{b}-\ln \mathrm{a}]_{\mathrm{a}}^{\mathrm{b}}$ $\phi=\frac{\mu_{\mathrm{o}} \mathrm{nIh}}{2 \pi}[\ln (\mathrm{b} / \mathrm{a})]$ $\because$ Self inductance $(L)=\frac{\mathrm{n} \phi}{\mathrm{I}}$ $\therefore \quad \mathrm{L}=\frac{\mu_{\mathrm{o}} \mathrm{n}^{2} \mathrm{~h}}{2 \pi} \ln (\mathrm{b} / \mathrm{a})$
TS- EAMCET-06.05.2019
Electro Magnetic Induction
154786
A coil is connected to an AC source with peak emf, $8 V$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The coil has resistance of $8 \Omega$. If the average power dissipated by the coil is $0.4 \mathrm{~W}$, then the inductance of the coil is
154787
The magnetic potential energy stored in a certain inductor is $25 \mathrm{~mJ}$, when the current in the inductor is $60 \mathrm{~mA}$. This inductor is of inductance
1 $1.389 \mathrm{H}$
2 $138.88 \mathrm{H}$
3 $0.138 \mathrm{H}$
4 $13.89 \mathrm{H}$
Explanation:
D Given, $\mathrm{U}=25 \mathrm{~mJ}=25 \times 10^{-3} \mathrm{~J}, \mathrm{I}=60 \mathrm{~mA}=$ $60 \times 10^{-3} \mathrm{~A}$ We know that, energy stored in inductor is, $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{~L}=\frac{2 \mathrm{U}}{\mathrm{I}^{2}}$ $\mathrm{~L}=\frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^{2}}$ $\mathrm{~L}=13.89 \mathrm{H}$
