154804
Two infinite long wires each carrying a current $10 \mathrm{~A}$ are bend to form a right angle as shown in the figure. Then the magnetic induction at ' $O$ ' is $\left[\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right]$
154806
The current in a coil of $L=40 \mathrm{mH}$ is to be increased uniformly from $1 \mathrm{~A}$ to $11 \mathrm{~A}$ in $4 \mathrm{milli}$ sec. The induced e.m.f. will be
1 $100 \mathrm{~V}$
2 $0.4 \mathrm{~V}$
3 $440 \mathrm{~V}$
4 $40 \mathrm{~V}$
Explanation:
A Given, Inductance $\mathrm{L}=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Change in current $=|\Delta \mathrm{I}|=1-11$ Time $\mathrm{t}=4 \times 10^{-3} \mathrm{~s}$ $=10 \mathrm{~A}$ $\because \quad$ Induced emf $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}}$ $\varepsilon=100 \mathrm{~V}$
VITEEE-2017
Electro Magnetic Induction
154807
A wire loop PQRSP formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in figure below. The magnitude of magnetic induction at centre $C$ is.
B Magnetic field at centre ' $\mathrm{C}$ ' of semicircle $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}$ Magnetic field due to semicircle of radius $\mathrm{R}_{1}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}$ Magnetic field due to semicircle of Radius $\mathrm{R}_{2}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ Net magnetic field $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}-\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
JCECE-2017
Electro Magnetic Induction
154808
Radii of two conducting circular loops are $b$ and a respectively, where $b>>$ a. Centre's of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops
B Magnetic field due to circle of radius b, $\mathrm{B}_{\mathrm{b}}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{~b}}$ Now, $\quad \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos 90^{\circ}$ $\phi=0$ $\phi=$ MI $\mathrm{M}=0$
154804
Two infinite long wires each carrying a current $10 \mathrm{~A}$ are bend to form a right angle as shown in the figure. Then the magnetic induction at ' $O$ ' is $\left[\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right]$
154806
The current in a coil of $L=40 \mathrm{mH}$ is to be increased uniformly from $1 \mathrm{~A}$ to $11 \mathrm{~A}$ in $4 \mathrm{milli}$ sec. The induced e.m.f. will be
1 $100 \mathrm{~V}$
2 $0.4 \mathrm{~V}$
3 $440 \mathrm{~V}$
4 $40 \mathrm{~V}$
Explanation:
A Given, Inductance $\mathrm{L}=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Change in current $=|\Delta \mathrm{I}|=1-11$ Time $\mathrm{t}=4 \times 10^{-3} \mathrm{~s}$ $=10 \mathrm{~A}$ $\because \quad$ Induced emf $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}}$ $\varepsilon=100 \mathrm{~V}$
VITEEE-2017
Electro Magnetic Induction
154807
A wire loop PQRSP formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in figure below. The magnitude of magnetic induction at centre $C$ is.
B Magnetic field at centre ' $\mathrm{C}$ ' of semicircle $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}$ Magnetic field due to semicircle of radius $\mathrm{R}_{1}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}$ Magnetic field due to semicircle of Radius $\mathrm{R}_{2}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ Net magnetic field $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}-\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
JCECE-2017
Electro Magnetic Induction
154808
Radii of two conducting circular loops are $b$ and a respectively, where $b>>$ a. Centre's of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops
B Magnetic field due to circle of radius b, $\mathrm{B}_{\mathrm{b}}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{~b}}$ Now, $\quad \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos 90^{\circ}$ $\phi=0$ $\phi=$ MI $\mathrm{M}=0$
154804
Two infinite long wires each carrying a current $10 \mathrm{~A}$ are bend to form a right angle as shown in the figure. Then the magnetic induction at ' $O$ ' is $\left[\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right]$
154806
The current in a coil of $L=40 \mathrm{mH}$ is to be increased uniformly from $1 \mathrm{~A}$ to $11 \mathrm{~A}$ in $4 \mathrm{milli}$ sec. The induced e.m.f. will be
1 $100 \mathrm{~V}$
2 $0.4 \mathrm{~V}$
3 $440 \mathrm{~V}$
4 $40 \mathrm{~V}$
Explanation:
A Given, Inductance $\mathrm{L}=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Change in current $=|\Delta \mathrm{I}|=1-11$ Time $\mathrm{t}=4 \times 10^{-3} \mathrm{~s}$ $=10 \mathrm{~A}$ $\because \quad$ Induced emf $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}}$ $\varepsilon=100 \mathrm{~V}$
VITEEE-2017
Electro Magnetic Induction
154807
A wire loop PQRSP formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in figure below. The magnitude of magnetic induction at centre $C$ is.
B Magnetic field at centre ' $\mathrm{C}$ ' of semicircle $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}$ Magnetic field due to semicircle of radius $\mathrm{R}_{1}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}$ Magnetic field due to semicircle of Radius $\mathrm{R}_{2}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ Net magnetic field $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}-\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
JCECE-2017
Electro Magnetic Induction
154808
Radii of two conducting circular loops are $b$ and a respectively, where $b>>$ a. Centre's of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops
B Magnetic field due to circle of radius b, $\mathrm{B}_{\mathrm{b}}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{~b}}$ Now, $\quad \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos 90^{\circ}$ $\phi=0$ $\phi=$ MI $\mathrm{M}=0$
154804
Two infinite long wires each carrying a current $10 \mathrm{~A}$ are bend to form a right angle as shown in the figure. Then the magnetic induction at ' $O$ ' is $\left[\mu_{0}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right]$
154806
The current in a coil of $L=40 \mathrm{mH}$ is to be increased uniformly from $1 \mathrm{~A}$ to $11 \mathrm{~A}$ in $4 \mathrm{milli}$ sec. The induced e.m.f. will be
1 $100 \mathrm{~V}$
2 $0.4 \mathrm{~V}$
3 $440 \mathrm{~V}$
4 $40 \mathrm{~V}$
Explanation:
A Given, Inductance $\mathrm{L}=40 \mathrm{mH}=40 \times 10^{-3} \mathrm{H}$ Change in current $=|\Delta \mathrm{I}|=1-11$ Time $\mathrm{t}=4 \times 10^{-3} \mathrm{~s}$ $=10 \mathrm{~A}$ $\because \quad$ Induced emf $\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}$ $\varepsilon=40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}}$ $\varepsilon=100 \mathrm{~V}$
VITEEE-2017
Electro Magnetic Induction
154807
A wire loop PQRSP formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in figure below. The magnitude of magnetic induction at centre $C$ is.
B Magnetic field at centre ' $\mathrm{C}$ ' of semicircle $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}}$ Magnetic field due to semicircle of radius $\mathrm{R}_{1}$ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}$ Magnetic field due to semicircle of Radius $\mathrm{R}_{2}$ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ Net magnetic field $\mathrm{B}_{\text {net }}=\mathrm{B}_{1}-\mathrm{B}_{2}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{1}}-\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}_{2}}$ $\mathrm{~B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{4}\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$
JCECE-2017
Electro Magnetic Induction
154808
Radii of two conducting circular loops are $b$ and a respectively, where $b>>$ a. Centre's of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops
B Magnetic field due to circle of radius b, $\mathrm{B}_{\mathrm{b}}=\frac{\mu_{\mathrm{o}} \mathrm{i}}{2 \mathrm{~b}}$ Now, $\quad \phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos 90^{\circ}$ $\phi=0$ $\phi=$ MI $\mathrm{M}=0$
