154656
Find the zero induced emf position for a conducting circular loop of radius $r$. The circular loop is rotated about its diameter at a constant angular velocity $\omega$ in magnetic field $B$ perpendicular to the axis of rotation.
1 $0^{\circ}$
2 $\frac{\pi}{3}$
3 $\frac{\pi}{4}$
4 $\frac{\pi}{2}$
Explanation:
A Given, radius of circular loop $=r$ Area of circular loop $(\mathrm{A})=\pi \mathrm{r}^{2}$ We know that, Flux $(\phi)=\mathrm{BA} \cos \theta$ $=\mathrm{B} \times \pi \mathrm{r}^{2} \times \cos \omega \mathrm{t} \quad(\because \theta=\omega \mathrm{t})$ Induced emf, $\varepsilon=\frac{d \phi}{d t}$ $\varepsilon=\frac{d\left(B \pi r^{2} \cos \omega t\right)}{d t}$ $\varepsilon=-B \pi r^{2} \omega t \sin \omega t$ $\operatorname{emf}(\varepsilon)=0$, if $\sin \omega \mathrm{t}=0$ $\text { or } \omega \mathrm{t}=0$ $\text { or } \theta=0^{\circ}$
TS EAMCET (Medical) 09.08.2021
Electro Magnetic Induction
154657
A long straight conductor carrying current $I$ and a square frame of side ' $a$ ' are in the same plane as shown in the figure. This frame moves with a constant velocity ' $v$ ' right side. The induced emf in the frame will be proportional to
1 $\frac{1}{(2 r-a)^{2}}$
2 $\frac{1}{(2 r+a)^{2}}$
3 $\frac{1}{(2 r-a)(2 r+a)}$
4 $\frac{1}{\mathrm{r}^{2}}$
Explanation:
C Given, current = I, sides of square = a velocity $=\mathrm{v}$ Then, Induced emf $(\varepsilon)=\mathrm{B}_{1} \mathrm{av}-\mathrm{B}_{2} \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\left(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}-\frac{\mathrm{a}}{2}\right)}-\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}+\frac{\mathrm{a}}{2}\right)}\right) \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{1}{r-\frac{a}{2}}-\frac{1}{r+\frac{a}{2}}\right)$ $q=\int_{0}^{5}(1.2 t+3) d t$ $\mathrm{q}=\left[\frac{1.2 \mathrm{t}^{2}}{2}+3 \mathrm{t}\right]_{0}^{5}$ $\mathrm{q}=\frac{1.2 \times 25}{2}+3 \times 5$ $\mathrm{q}=15+15$ $\mathrm{q}=30 \mathrm{C}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2}{(2 r-a)}-\frac{2}{(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2 r+a-2 r+a}{(2 r-a)(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \mathrm{Ia}^{2} \mathrm{v}}{\pi(2 \mathrm{r}-\mathrm{a})(2 \mathrm{r}+\mathrm{a})}$ So, $\quad \operatorname{emf}(\varepsilon) \propto \frac{1}{(2 r-a)(2 r+a)}$
UPSEE 2020
Electro Magnetic Induction
154658
A train is running at a speed of $72 \mathrm{~km} \mathrm{hr}^{-1}$ on the rails separated by a distance of $150 \mathrm{~cm}$. If the vertical component of earth's magnetic field at the place is $4.0 \times 10^{-5} \mathrm{~T}$. The induced emf on the rails is
1 $1.2 \mathrm{mV}$
2 $3 \mathrm{mV}$
3 $2.5 \mathrm{mV}$
4 $0.5 \mathrm{mV}$
5 $4.2 \mathrm{mV}$
Explanation:
A Given, Speed of train $(\mathrm{v})=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$ Distance $(l)=150 \mathrm{~cm}=1.5 \mathrm{~m}$ Earth's magnetic field $(\mathrm{B})=4 \times 10^{-5} \mathrm{~T}$. So, $\quad$ induced emf $(\varepsilon)=\mathrm{B} / \mathrm{v}$ $\operatorname{emf}(\varepsilon)=4 \times 10^{-5} \times 1.5 \times 20$ $\operatorname{emf}(\varepsilon)=1.2 \mathrm{mV}$
Kerala CEE 2020
Electro Magnetic Induction
154659
A source e.m.f.E $=15 \mathrm{~V}$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i=1.2 t+3$. Then the total charge that will flow in the first 5 sec will be
1 $10 \mathrm{C}$
2 $20 \mathrm{C}$
3 $30 \mathrm{C}$
4 $40 \mathrm{C}$
Explanation:
C Given, e.m.f. $\mathrm{E}=15 \mathrm{~V}$. Current (i) $=1.2 \mathrm{t}+3$ Time $(\mathrm{t})=5 \mathrm{sec}$ Charge $(\mathrm{q})=$ ? We know that, $\mathrm{dq}=\mathrm{idt}$ Integrating both side, $\int \mathrm{dq}=\int_{0}^{5} \mathrm{idt}$
154656
Find the zero induced emf position for a conducting circular loop of radius $r$. The circular loop is rotated about its diameter at a constant angular velocity $\omega$ in magnetic field $B$ perpendicular to the axis of rotation.
1 $0^{\circ}$
2 $\frac{\pi}{3}$
3 $\frac{\pi}{4}$
4 $\frac{\pi}{2}$
Explanation:
A Given, radius of circular loop $=r$ Area of circular loop $(\mathrm{A})=\pi \mathrm{r}^{2}$ We know that, Flux $(\phi)=\mathrm{BA} \cos \theta$ $=\mathrm{B} \times \pi \mathrm{r}^{2} \times \cos \omega \mathrm{t} \quad(\because \theta=\omega \mathrm{t})$ Induced emf, $\varepsilon=\frac{d \phi}{d t}$ $\varepsilon=\frac{d\left(B \pi r^{2} \cos \omega t\right)}{d t}$ $\varepsilon=-B \pi r^{2} \omega t \sin \omega t$ $\operatorname{emf}(\varepsilon)=0$, if $\sin \omega \mathrm{t}=0$ $\text { or } \omega \mathrm{t}=0$ $\text { or } \theta=0^{\circ}$
TS EAMCET (Medical) 09.08.2021
Electro Magnetic Induction
154657
A long straight conductor carrying current $I$ and a square frame of side ' $a$ ' are in the same plane as shown in the figure. This frame moves with a constant velocity ' $v$ ' right side. The induced emf in the frame will be proportional to
1 $\frac{1}{(2 r-a)^{2}}$
2 $\frac{1}{(2 r+a)^{2}}$
3 $\frac{1}{(2 r-a)(2 r+a)}$
4 $\frac{1}{\mathrm{r}^{2}}$
Explanation:
C Given, current = I, sides of square = a velocity $=\mathrm{v}$ Then, Induced emf $(\varepsilon)=\mathrm{B}_{1} \mathrm{av}-\mathrm{B}_{2} \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\left(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}-\frac{\mathrm{a}}{2}\right)}-\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}+\frac{\mathrm{a}}{2}\right)}\right) \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{1}{r-\frac{a}{2}}-\frac{1}{r+\frac{a}{2}}\right)$ $q=\int_{0}^{5}(1.2 t+3) d t$ $\mathrm{q}=\left[\frac{1.2 \mathrm{t}^{2}}{2}+3 \mathrm{t}\right]_{0}^{5}$ $\mathrm{q}=\frac{1.2 \times 25}{2}+3 \times 5$ $\mathrm{q}=15+15$ $\mathrm{q}=30 \mathrm{C}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2}{(2 r-a)}-\frac{2}{(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2 r+a-2 r+a}{(2 r-a)(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \mathrm{Ia}^{2} \mathrm{v}}{\pi(2 \mathrm{r}-\mathrm{a})(2 \mathrm{r}+\mathrm{a})}$ So, $\quad \operatorname{emf}(\varepsilon) \propto \frac{1}{(2 r-a)(2 r+a)}$
UPSEE 2020
Electro Magnetic Induction
154658
A train is running at a speed of $72 \mathrm{~km} \mathrm{hr}^{-1}$ on the rails separated by a distance of $150 \mathrm{~cm}$. If the vertical component of earth's magnetic field at the place is $4.0 \times 10^{-5} \mathrm{~T}$. The induced emf on the rails is
1 $1.2 \mathrm{mV}$
2 $3 \mathrm{mV}$
3 $2.5 \mathrm{mV}$
4 $0.5 \mathrm{mV}$
5 $4.2 \mathrm{mV}$
Explanation:
A Given, Speed of train $(\mathrm{v})=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$ Distance $(l)=150 \mathrm{~cm}=1.5 \mathrm{~m}$ Earth's magnetic field $(\mathrm{B})=4 \times 10^{-5} \mathrm{~T}$. So, $\quad$ induced emf $(\varepsilon)=\mathrm{B} / \mathrm{v}$ $\operatorname{emf}(\varepsilon)=4 \times 10^{-5} \times 1.5 \times 20$ $\operatorname{emf}(\varepsilon)=1.2 \mathrm{mV}$
Kerala CEE 2020
Electro Magnetic Induction
154659
A source e.m.f.E $=15 \mathrm{~V}$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i=1.2 t+3$. Then the total charge that will flow in the first 5 sec will be
1 $10 \mathrm{C}$
2 $20 \mathrm{C}$
3 $30 \mathrm{C}$
4 $40 \mathrm{C}$
Explanation:
C Given, e.m.f. $\mathrm{E}=15 \mathrm{~V}$. Current (i) $=1.2 \mathrm{t}+3$ Time $(\mathrm{t})=5 \mathrm{sec}$ Charge $(\mathrm{q})=$ ? We know that, $\mathrm{dq}=\mathrm{idt}$ Integrating both side, $\int \mathrm{dq}=\int_{0}^{5} \mathrm{idt}$
154656
Find the zero induced emf position for a conducting circular loop of radius $r$. The circular loop is rotated about its diameter at a constant angular velocity $\omega$ in magnetic field $B$ perpendicular to the axis of rotation.
1 $0^{\circ}$
2 $\frac{\pi}{3}$
3 $\frac{\pi}{4}$
4 $\frac{\pi}{2}$
Explanation:
A Given, radius of circular loop $=r$ Area of circular loop $(\mathrm{A})=\pi \mathrm{r}^{2}$ We know that, Flux $(\phi)=\mathrm{BA} \cos \theta$ $=\mathrm{B} \times \pi \mathrm{r}^{2} \times \cos \omega \mathrm{t} \quad(\because \theta=\omega \mathrm{t})$ Induced emf, $\varepsilon=\frac{d \phi}{d t}$ $\varepsilon=\frac{d\left(B \pi r^{2} \cos \omega t\right)}{d t}$ $\varepsilon=-B \pi r^{2} \omega t \sin \omega t$ $\operatorname{emf}(\varepsilon)=0$, if $\sin \omega \mathrm{t}=0$ $\text { or } \omega \mathrm{t}=0$ $\text { or } \theta=0^{\circ}$
TS EAMCET (Medical) 09.08.2021
Electro Magnetic Induction
154657
A long straight conductor carrying current $I$ and a square frame of side ' $a$ ' are in the same plane as shown in the figure. This frame moves with a constant velocity ' $v$ ' right side. The induced emf in the frame will be proportional to
1 $\frac{1}{(2 r-a)^{2}}$
2 $\frac{1}{(2 r+a)^{2}}$
3 $\frac{1}{(2 r-a)(2 r+a)}$
4 $\frac{1}{\mathrm{r}^{2}}$
Explanation:
C Given, current = I, sides of square = a velocity $=\mathrm{v}$ Then, Induced emf $(\varepsilon)=\mathrm{B}_{1} \mathrm{av}-\mathrm{B}_{2} \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\left(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}-\frac{\mathrm{a}}{2}\right)}-\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}+\frac{\mathrm{a}}{2}\right)}\right) \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{1}{r-\frac{a}{2}}-\frac{1}{r+\frac{a}{2}}\right)$ $q=\int_{0}^{5}(1.2 t+3) d t$ $\mathrm{q}=\left[\frac{1.2 \mathrm{t}^{2}}{2}+3 \mathrm{t}\right]_{0}^{5}$ $\mathrm{q}=\frac{1.2 \times 25}{2}+3 \times 5$ $\mathrm{q}=15+15$ $\mathrm{q}=30 \mathrm{C}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2}{(2 r-a)}-\frac{2}{(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2 r+a-2 r+a}{(2 r-a)(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \mathrm{Ia}^{2} \mathrm{v}}{\pi(2 \mathrm{r}-\mathrm{a})(2 \mathrm{r}+\mathrm{a})}$ So, $\quad \operatorname{emf}(\varepsilon) \propto \frac{1}{(2 r-a)(2 r+a)}$
UPSEE 2020
Electro Magnetic Induction
154658
A train is running at a speed of $72 \mathrm{~km} \mathrm{hr}^{-1}$ on the rails separated by a distance of $150 \mathrm{~cm}$. If the vertical component of earth's magnetic field at the place is $4.0 \times 10^{-5} \mathrm{~T}$. The induced emf on the rails is
1 $1.2 \mathrm{mV}$
2 $3 \mathrm{mV}$
3 $2.5 \mathrm{mV}$
4 $0.5 \mathrm{mV}$
5 $4.2 \mathrm{mV}$
Explanation:
A Given, Speed of train $(\mathrm{v})=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$ Distance $(l)=150 \mathrm{~cm}=1.5 \mathrm{~m}$ Earth's magnetic field $(\mathrm{B})=4 \times 10^{-5} \mathrm{~T}$. So, $\quad$ induced emf $(\varepsilon)=\mathrm{B} / \mathrm{v}$ $\operatorname{emf}(\varepsilon)=4 \times 10^{-5} \times 1.5 \times 20$ $\operatorname{emf}(\varepsilon)=1.2 \mathrm{mV}$
Kerala CEE 2020
Electro Magnetic Induction
154659
A source e.m.f.E $=15 \mathrm{~V}$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i=1.2 t+3$. Then the total charge that will flow in the first 5 sec will be
1 $10 \mathrm{C}$
2 $20 \mathrm{C}$
3 $30 \mathrm{C}$
4 $40 \mathrm{C}$
Explanation:
C Given, e.m.f. $\mathrm{E}=15 \mathrm{~V}$. Current (i) $=1.2 \mathrm{t}+3$ Time $(\mathrm{t})=5 \mathrm{sec}$ Charge $(\mathrm{q})=$ ? We know that, $\mathrm{dq}=\mathrm{idt}$ Integrating both side, $\int \mathrm{dq}=\int_{0}^{5} \mathrm{idt}$
154656
Find the zero induced emf position for a conducting circular loop of radius $r$. The circular loop is rotated about its diameter at a constant angular velocity $\omega$ in magnetic field $B$ perpendicular to the axis of rotation.
1 $0^{\circ}$
2 $\frac{\pi}{3}$
3 $\frac{\pi}{4}$
4 $\frac{\pi}{2}$
Explanation:
A Given, radius of circular loop $=r$ Area of circular loop $(\mathrm{A})=\pi \mathrm{r}^{2}$ We know that, Flux $(\phi)=\mathrm{BA} \cos \theta$ $=\mathrm{B} \times \pi \mathrm{r}^{2} \times \cos \omega \mathrm{t} \quad(\because \theta=\omega \mathrm{t})$ Induced emf, $\varepsilon=\frac{d \phi}{d t}$ $\varepsilon=\frac{d\left(B \pi r^{2} \cos \omega t\right)}{d t}$ $\varepsilon=-B \pi r^{2} \omega t \sin \omega t$ $\operatorname{emf}(\varepsilon)=0$, if $\sin \omega \mathrm{t}=0$ $\text { or } \omega \mathrm{t}=0$ $\text { or } \theta=0^{\circ}$
TS EAMCET (Medical) 09.08.2021
Electro Magnetic Induction
154657
A long straight conductor carrying current $I$ and a square frame of side ' $a$ ' are in the same plane as shown in the figure. This frame moves with a constant velocity ' $v$ ' right side. The induced emf in the frame will be proportional to
1 $\frac{1}{(2 r-a)^{2}}$
2 $\frac{1}{(2 r+a)^{2}}$
3 $\frac{1}{(2 r-a)(2 r+a)}$
4 $\frac{1}{\mathrm{r}^{2}}$
Explanation:
C Given, current = I, sides of square = a velocity $=\mathrm{v}$ Then, Induced emf $(\varepsilon)=\mathrm{B}_{1} \mathrm{av}-\mathrm{B}_{2} \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\left(\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}-\frac{\mathrm{a}}{2}\right)}-\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi\left(\mathrm{r}+\frac{\mathrm{a}}{2}\right)}\right) \mathrm{av}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{1}{r-\frac{a}{2}}-\frac{1}{r+\frac{a}{2}}\right)$ $q=\int_{0}^{5}(1.2 t+3) d t$ $\mathrm{q}=\left[\frac{1.2 \mathrm{t}^{2}}{2}+3 \mathrm{t}\right]_{0}^{5}$ $\mathrm{q}=\frac{1.2 \times 25}{2}+3 \times 5$ $\mathrm{q}=15+15$ $\mathrm{q}=30 \mathrm{C}$ $\operatorname{emf}(\varepsilon)=\frac{\mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2}{(2 r-a)}-\frac{2}{(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \operatorname{Iav}}{2 \pi}\left(\frac{2 r+a-2 r+a}{(2 r-a)(2 r+a)}\right)$ $\operatorname{emf}(\varepsilon)=\frac{2 \mu_{0} \mathrm{Ia}^{2} \mathrm{v}}{\pi(2 \mathrm{r}-\mathrm{a})(2 \mathrm{r}+\mathrm{a})}$ So, $\quad \operatorname{emf}(\varepsilon) \propto \frac{1}{(2 r-a)(2 r+a)}$
UPSEE 2020
Electro Magnetic Induction
154658
A train is running at a speed of $72 \mathrm{~km} \mathrm{hr}^{-1}$ on the rails separated by a distance of $150 \mathrm{~cm}$. If the vertical component of earth's magnetic field at the place is $4.0 \times 10^{-5} \mathrm{~T}$. The induced emf on the rails is
1 $1.2 \mathrm{mV}$
2 $3 \mathrm{mV}$
3 $2.5 \mathrm{mV}$
4 $0.5 \mathrm{mV}$
5 $4.2 \mathrm{mV}$
Explanation:
A Given, Speed of train $(\mathrm{v})=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$ Distance $(l)=150 \mathrm{~cm}=1.5 \mathrm{~m}$ Earth's magnetic field $(\mathrm{B})=4 \times 10^{-5} \mathrm{~T}$. So, $\quad$ induced emf $(\varepsilon)=\mathrm{B} / \mathrm{v}$ $\operatorname{emf}(\varepsilon)=4 \times 10^{-5} \times 1.5 \times 20$ $\operatorname{emf}(\varepsilon)=1.2 \mathrm{mV}$
Kerala CEE 2020
Electro Magnetic Induction
154659
A source e.m.f.E $=15 \mathrm{~V}$ and having negligible internal resistance is connected to a variable resistance so that the current in the circuit increases with time as $i=1.2 t+3$. Then the total charge that will flow in the first 5 sec will be
1 $10 \mathrm{C}$
2 $20 \mathrm{C}$
3 $30 \mathrm{C}$
4 $40 \mathrm{C}$
Explanation:
C Given, e.m.f. $\mathrm{E}=15 \mathrm{~V}$. Current (i) $=1.2 \mathrm{t}+3$ Time $(\mathrm{t})=5 \mathrm{sec}$ Charge $(\mathrm{q})=$ ? We know that, $\mathrm{dq}=\mathrm{idt}$ Integrating both side, $\int \mathrm{dq}=\int_{0}^{5} \mathrm{idt}$
