B Magnetic field due to straight wire of infinite length is, $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ If the point is on a line perpendicular to its length while at the centre of a semicircular coil $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{a}}+\mathrm{B}_{\mathrm{b}}+\mathrm{B}_{\mathrm{c}}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\pi \mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}$ $\mathrm{B}_{\text {net }}=\left(\frac{\mu_{\mathrm{o}}}{4 \pi}\right)\left(\frac{\mathrm{i}}{\mathrm{r}}\right)(\pi+2)$
VITEEE-2012
Electro Magnetic Induction
154580
A coil of resistance $10 \Omega$ and inductance $5 \mathrm{H}$ is connected to a $100 \mathrm{~V}$ battery. Then the energy stored in the coil is
1 $250 \mathrm{~J}$
2 $250 \mathrm{erg}$
3 $125 \mathrm{~J}$
4 $125 \mathrm{erg}$
Explanation:
A Given that, $\mathrm{R}=10 \Omega$ $\mathrm{L}=5 \mathrm{H}$ $\mathrm{V}=100 \mathrm{~V}$ $\mathrm{H}=?$ Steady state current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{100}{10}=10 \mathrm{Amp}$ Energy stored in the coil $(\mathrm{U})=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 5 \times 10 \times 10$ $\mathrm{U}=\frac{500}{2}=250 \mathrm{~J}$
VITEEE-2011
Electro Magnetic Induction
154581
In a magnetic field of $0.05 \mathrm{~T}$, area of a coil changes from $101 \mathrm{~cm}^{2}$ to $100 \mathrm{~cm}^{2}$ without changing the resistance which is $2 \Omega$. The amount of charge that flow during this period is
1 $2.5 \times 10^{-6} \mathrm{C}$
2 $2 \times 10^{-6} \mathrm{C}$
3 $10 \times 10^{-6} \mathrm{C}$
4 $8 \times 10^{-6} \mathrm{C}$
Explanation:
A Given that, $\mathrm{B}=0.05 \mathrm{~T}$ $\mathrm{R}=2 \Omega$ $\Delta \mathrm{A}=\mathrm{A}_{1}-\mathrm{A}_{2}=101-100=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ We know that, Change in flux $(\Delta \phi)=$ B. $\Delta \mathrm{A}=0.05 \times 10^{-4}$ $=5 \times 10^{-6} \mathrm{~Wb}$ Charge flowing $(\mathrm{q})=\frac{\Delta \phi}{\mathrm{R}}=\frac{5 \times 10^{-6}}{2}=2.5 \times 10^{-6} \mathrm{C}$
VITEEE-2011
Electro Magnetic Induction
154582
A wheel with 10 spokes each of length $L \mathbf{m}$ is rotated with a uniform angular velocity $\omega$ in a plane normal to the magnetic field $B$. The emf induced between the axle and the rim of the wheel is
B Given that, Number of spoke $(\mathrm{N})=10$ Length $=\mathrm{L} \mathrm{m}$ Angular velocity $=\omega$ Magnetic field $=\mathrm{B}$ Emf induced between the axle and rim, $\varepsilon=\frac{\mathrm{B} \omega \mathrm{L}^{2}}{2}$
B Magnetic field due to straight wire of infinite length is, $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ If the point is on a line perpendicular to its length while at the centre of a semicircular coil $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{a}}+\mathrm{B}_{\mathrm{b}}+\mathrm{B}_{\mathrm{c}}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\pi \mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}$ $\mathrm{B}_{\text {net }}=\left(\frac{\mu_{\mathrm{o}}}{4 \pi}\right)\left(\frac{\mathrm{i}}{\mathrm{r}}\right)(\pi+2)$
VITEEE-2012
Electro Magnetic Induction
154580
A coil of resistance $10 \Omega$ and inductance $5 \mathrm{H}$ is connected to a $100 \mathrm{~V}$ battery. Then the energy stored in the coil is
1 $250 \mathrm{~J}$
2 $250 \mathrm{erg}$
3 $125 \mathrm{~J}$
4 $125 \mathrm{erg}$
Explanation:
A Given that, $\mathrm{R}=10 \Omega$ $\mathrm{L}=5 \mathrm{H}$ $\mathrm{V}=100 \mathrm{~V}$ $\mathrm{H}=?$ Steady state current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{100}{10}=10 \mathrm{Amp}$ Energy stored in the coil $(\mathrm{U})=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 5 \times 10 \times 10$ $\mathrm{U}=\frac{500}{2}=250 \mathrm{~J}$
VITEEE-2011
Electro Magnetic Induction
154581
In a magnetic field of $0.05 \mathrm{~T}$, area of a coil changes from $101 \mathrm{~cm}^{2}$ to $100 \mathrm{~cm}^{2}$ without changing the resistance which is $2 \Omega$. The amount of charge that flow during this period is
1 $2.5 \times 10^{-6} \mathrm{C}$
2 $2 \times 10^{-6} \mathrm{C}$
3 $10 \times 10^{-6} \mathrm{C}$
4 $8 \times 10^{-6} \mathrm{C}$
Explanation:
A Given that, $\mathrm{B}=0.05 \mathrm{~T}$ $\mathrm{R}=2 \Omega$ $\Delta \mathrm{A}=\mathrm{A}_{1}-\mathrm{A}_{2}=101-100=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ We know that, Change in flux $(\Delta \phi)=$ B. $\Delta \mathrm{A}=0.05 \times 10^{-4}$ $=5 \times 10^{-6} \mathrm{~Wb}$ Charge flowing $(\mathrm{q})=\frac{\Delta \phi}{\mathrm{R}}=\frac{5 \times 10^{-6}}{2}=2.5 \times 10^{-6} \mathrm{C}$
VITEEE-2011
Electro Magnetic Induction
154582
A wheel with 10 spokes each of length $L \mathbf{m}$ is rotated with a uniform angular velocity $\omega$ in a plane normal to the magnetic field $B$. The emf induced between the axle and the rim of the wheel is
B Given that, Number of spoke $(\mathrm{N})=10$ Length $=\mathrm{L} \mathrm{m}$ Angular velocity $=\omega$ Magnetic field $=\mathrm{B}$ Emf induced between the axle and rim, $\varepsilon=\frac{\mathrm{B} \omega \mathrm{L}^{2}}{2}$
B Magnetic field due to straight wire of infinite length is, $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ If the point is on a line perpendicular to its length while at the centre of a semicircular coil $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{a}}+\mathrm{B}_{\mathrm{b}}+\mathrm{B}_{\mathrm{c}}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\pi \mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}$ $\mathrm{B}_{\text {net }}=\left(\frac{\mu_{\mathrm{o}}}{4 \pi}\right)\left(\frac{\mathrm{i}}{\mathrm{r}}\right)(\pi+2)$
VITEEE-2012
Electro Magnetic Induction
154580
A coil of resistance $10 \Omega$ and inductance $5 \mathrm{H}$ is connected to a $100 \mathrm{~V}$ battery. Then the energy stored in the coil is
1 $250 \mathrm{~J}$
2 $250 \mathrm{erg}$
3 $125 \mathrm{~J}$
4 $125 \mathrm{erg}$
Explanation:
A Given that, $\mathrm{R}=10 \Omega$ $\mathrm{L}=5 \mathrm{H}$ $\mathrm{V}=100 \mathrm{~V}$ $\mathrm{H}=?$ Steady state current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{100}{10}=10 \mathrm{Amp}$ Energy stored in the coil $(\mathrm{U})=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 5 \times 10 \times 10$ $\mathrm{U}=\frac{500}{2}=250 \mathrm{~J}$
VITEEE-2011
Electro Magnetic Induction
154581
In a magnetic field of $0.05 \mathrm{~T}$, area of a coil changes from $101 \mathrm{~cm}^{2}$ to $100 \mathrm{~cm}^{2}$ without changing the resistance which is $2 \Omega$. The amount of charge that flow during this period is
1 $2.5 \times 10^{-6} \mathrm{C}$
2 $2 \times 10^{-6} \mathrm{C}$
3 $10 \times 10^{-6} \mathrm{C}$
4 $8 \times 10^{-6} \mathrm{C}$
Explanation:
A Given that, $\mathrm{B}=0.05 \mathrm{~T}$ $\mathrm{R}=2 \Omega$ $\Delta \mathrm{A}=\mathrm{A}_{1}-\mathrm{A}_{2}=101-100=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ We know that, Change in flux $(\Delta \phi)=$ B. $\Delta \mathrm{A}=0.05 \times 10^{-4}$ $=5 \times 10^{-6} \mathrm{~Wb}$ Charge flowing $(\mathrm{q})=\frac{\Delta \phi}{\mathrm{R}}=\frac{5 \times 10^{-6}}{2}=2.5 \times 10^{-6} \mathrm{C}$
VITEEE-2011
Electro Magnetic Induction
154582
A wheel with 10 spokes each of length $L \mathbf{m}$ is rotated with a uniform angular velocity $\omega$ in a plane normal to the magnetic field $B$. The emf induced between the axle and the rim of the wheel is
B Given that, Number of spoke $(\mathrm{N})=10$ Length $=\mathrm{L} \mathrm{m}$ Angular velocity $=\omega$ Magnetic field $=\mathrm{B}$ Emf induced between the axle and rim, $\varepsilon=\frac{\mathrm{B} \omega \mathrm{L}^{2}}{2}$
B Magnetic field due to straight wire of infinite length is, $\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{r}}$ If the point is on a line perpendicular to its length while at the centre of a semicircular coil $=\frac{\mu_{0} \pi \mathrm{i}}{4 \pi \mathrm{r}}$ $\therefore \quad \mathrm{B}_{\text {net }}=\mathrm{B}_{\mathrm{a}}+\mathrm{B}_{\mathrm{b}}+\mathrm{B}_{\mathrm{c}}$ $\mathrm{B}_{\text {net }}=\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\pi \mathrm{i}}{\mathrm{r}}+\frac{\mu_{\mathrm{o}}}{4 \pi} \frac{\mathrm{i}}{\mathrm{r}}$ $\mathrm{B}_{\text {net }}=\left(\frac{\mu_{\mathrm{o}}}{4 \pi}\right)\left(\frac{\mathrm{i}}{\mathrm{r}}\right)(\pi+2)$
VITEEE-2012
Electro Magnetic Induction
154580
A coil of resistance $10 \Omega$ and inductance $5 \mathrm{H}$ is connected to a $100 \mathrm{~V}$ battery. Then the energy stored in the coil is
1 $250 \mathrm{~J}$
2 $250 \mathrm{erg}$
3 $125 \mathrm{~J}$
4 $125 \mathrm{erg}$
Explanation:
A Given that, $\mathrm{R}=10 \Omega$ $\mathrm{L}=5 \mathrm{H}$ $\mathrm{V}=100 \mathrm{~V}$ $\mathrm{H}=?$ Steady state current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{100}{10}=10 \mathrm{Amp}$ Energy stored in the coil $(\mathrm{U})=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 5 \times 10 \times 10$ $\mathrm{U}=\frac{500}{2}=250 \mathrm{~J}$
VITEEE-2011
Electro Magnetic Induction
154581
In a magnetic field of $0.05 \mathrm{~T}$, area of a coil changes from $101 \mathrm{~cm}^{2}$ to $100 \mathrm{~cm}^{2}$ without changing the resistance which is $2 \Omega$. The amount of charge that flow during this period is
1 $2.5 \times 10^{-6} \mathrm{C}$
2 $2 \times 10^{-6} \mathrm{C}$
3 $10 \times 10^{-6} \mathrm{C}$
4 $8 \times 10^{-6} \mathrm{C}$
Explanation:
A Given that, $\mathrm{B}=0.05 \mathrm{~T}$ $\mathrm{R}=2 \Omega$ $\Delta \mathrm{A}=\mathrm{A}_{1}-\mathrm{A}_{2}=101-100=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ We know that, Change in flux $(\Delta \phi)=$ B. $\Delta \mathrm{A}=0.05 \times 10^{-4}$ $=5 \times 10^{-6} \mathrm{~Wb}$ Charge flowing $(\mathrm{q})=\frac{\Delta \phi}{\mathrm{R}}=\frac{5 \times 10^{-6}}{2}=2.5 \times 10^{-6} \mathrm{C}$
VITEEE-2011
Electro Magnetic Induction
154582
A wheel with 10 spokes each of length $L \mathbf{m}$ is rotated with a uniform angular velocity $\omega$ in a plane normal to the magnetic field $B$. The emf induced between the axle and the rim of the wheel is
B Given that, Number of spoke $(\mathrm{N})=10$ Length $=\mathrm{L} \mathrm{m}$ Angular velocity $=\omega$ Magnetic field $=\mathrm{B}$ Emf induced between the axle and rim, $\varepsilon=\frac{\mathrm{B} \omega \mathrm{L}^{2}}{2}$
