MCQ 11 Mark
What is the energy required to build up a current of 1 A in an inductor of 20 mH?
- ✓10 mJ
- B20 mJ
- C20 J
- D10 J
Answer
View full question & answer→Correct option: A.
10 mJ
10 mJ
Questions
M.C.Q (1 Marks)
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32 questions · auto-graded multiple-choice test.
MCQ 21 Mark
A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
- ✓0.1 V
- B1 V
- C10 V
- D0.01 V
Answer
View full question & answer→Correct option: A.
0.1 V
0.1 V
MCQ 31 Mark
Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
- ✓20 mH
- B30 mH
- C10 mH
- D$\frac{20}{3} mH$
Answer
View full question & answer→Correct option: A.
20 mH
20 mH
MCQ 41 Mark
A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
- ✓BLv
- B$BLV ^2$
- C$\frac{1}{2}$ Blv
- D$\frac{2 Bl }{ v }$
Answer
View full question & answer→Correct option: A.
BLv
BLv
MCQ 51 Mark
A circular coil of 100 turns with a cross-sectional area (A) of 1 m2 is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
- A1 Wb
- ✓100 Wb
- C50 Wb
- D200 Wb
Answer
View full question & answer→Correct option: B.
100 Wb
100 Wb
MCQ 61 Mark
The mutual inductance $(M)$ of the two coils is $3 H$. The self inductances of the coils are $4 H$ and $9 H$ respectively. The coefficient of coupling between the coils is
- A0.3
- B0.4
- ✓0.5
- D0.6
Answer
View full question & answer→Correct option: C.
0.5
(c): Given, $M=3 H , L_1=4 H , L_2=9 H$
$M=K \sqrt{L_1 L_2} ; 3=K \sqrt{4 \times 9} ; K=0.5$
$M=K \sqrt{L_1 L_2} ; 3=K \sqrt{4 \times 9} ; K=0.5$
MCQ 71 Mark
If current ' $l$ ' is flowing in the closed circuit with collective resistance ' $R$ ', the rate of production of heat energy in the loop as we pull it along with a constant speed ' $V$ ' is ( $L=$ length of conductor; $B=$ magnetic field)
- A$\frac{B L V}{R}$
- B$\frac{B^2 L^2 V^2}{R^2}$
- C$\frac{B L V}{R^2}$
- ✓$\frac{B^2 L^2 V^2}{R}$
Answer
View full question & answer→Correct option: D.
$\frac{B^2 L^2 V^2}{R}$
(d) : Motional emf, $e=B V L$
Heat generated $=\frac{e^2}{R}=\frac{B^2 L^2 V^2}{R} \quad$
Heat generated $=\frac{e^2}{R}=\frac{B^2 L^2 V^2}{R} \quad$
MCQ 81 Mark
Two coils $A$ and $B$ have mutual inductance $0.008 H$. The current changes in the coil $A$, according to the equation $I=I_m \sin \omega t_{\text {, }}$ where $I_m=5 A$ and $0=200 \pi rad s ^{-1}$. The maximum value of the e.m.f. induced in the coil $B$ in volt is
- A$4 \pi$
- ✓$8 \pi$
- C$10 \pi$
- D$16 \pi$
Answer
View full question & answer→Correct option: B.
$8 \pi$
(b) : $M=0.008 H , I=I_m \sin \theta t, I_m=5 A$ and
$\omega=200 \pi rad / s$
Induced emf, $e=M \frac{d I}{d t} ; \quad \frac{d I}{d t}=I_m \times \omega \cos \omega t$
$\therefore \quad e=M I_m \omega \cos \omega t$
For maximum value, $\cos \varphi t=1$
$
\therefore e=0.008 \times 5 \times 200 \pi \times 1=8 \pi V
$
$\omega=200 \pi rad / s$
Induced emf, $e=M \frac{d I}{d t} ; \quad \frac{d I}{d t}=I_m \times \omega \cos \omega t$
$\therefore \quad e=M I_m \omega \cos \omega t$
For maximum value, $\cos \varphi t=1$
$
\therefore e=0.008 \times 5 \times 200 \pi \times 1=8 \pi V
$
MCQ 91 Mark
The magnetic flux through a loop of resistance $10 \Omega$ is varying according to the relation $\phi=6 t^2+7 t+1$, where $\phi$ is in milliweber, time is in second then at time $t=1$ s the induced e.m.f. is
- A$12 mV$
- B$7 mV$
- ✓$19 mV$
- D$19 V$
Answer
View full question & answer→Correct option: C.
$19 mV$
(c) : Given, $R=10 \Omega, \phi=6 t^2+7 t+1, t=1 s$
$
\begin{aligned}
& \varepsilon=\frac{d \phi}{d t} ; \varepsilon=\frac{d}{d t}\left(6 t^2+7 t+1\right)=12 t+7 \\
& \varepsilon=12 \times 1+7=19 mV
\end{aligned}
$
$
\begin{aligned}
& \varepsilon=\frac{d \phi}{d t} ; \varepsilon=\frac{d}{d t}\left(6 t^2+7 t+1\right)=12 t+7 \\
& \varepsilon=12 \times 1+7=19 mV
\end{aligned}
$
MCQ 101 Mark
Self inductance of solenoid is
- Adirectly proportional to current flowing through the coil.
- Bdirectly proportional to the length.
- ✓directly proportional to its area of crosssection.
- Dinversely proportional to the area of crosssection.
Answer
View full question & answer→Correct option: C.
directly proportional to its area of crosssection.
(c) : As, $L=\frac{\mu_0 N^2 A}{l} ; L \propto A$
MCQ 111 Mark
A conducting wire of length $2500 m$ is kept in east-west direction, at a height of $10 m$ from the ground. If it falls freely on the ground then the current induced in the wire is (Resistance of wire $=25 \sqrt{2} \Omega$, acceleration due to gravity $g=10 m / s ^2$, $B_H=2 \times 10^{-5} T$ )
- A$0.2 A$
- ✓$0.02 A$
- C$0.01 A$
- D$2 A$
Answer
View full question & answer→Correct option: B.
$0.02 A$
(b) : Given, $l=2500 m , H=10 m , R=25 \sqrt{2} \Omega$
$
\begin{aligned}
& g=10 m / s ^2, B_H=2 \times 10^{-5} T \\
& v=\sqrt{2 g H}=\sqrt{2 \times 10 \times 10}=10 \sqrt{2} m / s \\
& e=B v l=2 \times 10^{-5} \times 10 \sqrt{2} \times 2500 \\
& i=\frac{e}{R}=\frac{2 \times 10^{-5} \times 10 \sqrt{2} \times 2500}{25 \sqrt{2}} ; i=0.02 A
\end{aligned}
$
$
\begin{aligned}
& g=10 m / s ^2, B_H=2 \times 10^{-5} T \\
& v=\sqrt{2 g H}=\sqrt{2 \times 10 \times 10}=10 \sqrt{2} m / s \\
& e=B v l=2 \times 10^{-5} \times 10 \sqrt{2} \times 2500 \\
& i=\frac{e}{R}=\frac{2 \times 10^{-5} \times 10 \sqrt{2} \times 2500}{25 \sqrt{2}} ; i=0.02 A
\end{aligned}
$
MCQ 121 Mark
The selfinduction ( $L$ ) produced by solenoid oflength ' $l$' having ' $N$ number of turns and cross sectional area ' $A$ ' is given by the formula ( $\phi=$ magnetic flux, $\mu_0=$ permeability of vacuum)
- A$L=N \phi$
- B$L=\mu_0 NAl$
- ✓$L=\frac{\mu_0 N^2 A}{l}$
- D$L=\frac{\mu_0 N A}{l}$
Answer
View full question & answer→Correct option: C.
$L=\frac{\mu_0 N^2 A}{l}$
(c) : $L=\frac{\mu_0 N^2 A}{l}$
MCQ 131 Mark
A long solenoid has 1500 turns. When a current of 3.5 A flows through it, the magnetic flux linked with each turn of solenoid is $2.8 \times 10^{-3}$ weber. The self-inductance of solenoid is
- ✓$1.2 H$
- B$2.4 H$
- C$3.6 H$
- D$6 H$
Answer
View full question & answer→Correct option: A.
$1.2 H$
(a) : $N=1500, i=3.5 A , \phi=2.8 \times 10^{-3} Wb$
As, $N \phi=L I$
$
L=\frac{2.8 \times 10^{-3} \times 1500}{3.5} ; L=\frac{4}{5} \times 1500 \times 10^{-3}=1.2 H
$
As, $N \phi=L I$
$
L=\frac{2.8 \times 10^{-3} \times 1500}{3.5} ; L=\frac{4}{5} \times 1500 \times 10^{-3}=1.2 H
$
MCQ 141 Mark
A coil having effective area $A$, is held with its plane normal to magnetic field of induction $B$. The magnetic induction is quickly reduced by $25 \%$ of its initial value in 2 second. Then the e.m.f. induced across the coil will be
- A$\frac{A B}{4}$
- B$\frac{A B}{2}$
- C$\frac{3 A B}{4}$
- ✓$\frac{3 A B}{8}$
Answer
View full question & answer→Correct option: D.
$\frac{3 A B}{8}$
(d) : As, $e=-\frac{d \phi}{d t}=-\frac{d}{d t} B A$,
$
B^{\prime}=\frac{100 B-25 B}{100}=\frac{3}{4} B ;|e|=\left|-A \frac{d B}{d t}\right| ; e=A \frac{\frac{3}{4} \times B}{2}=\frac{3 A B}{8}
$
$
B^{\prime}=\frac{100 B-25 B}{100}=\frac{3}{4} B ;|e|=\left|-A \frac{d B}{d t}\right| ; e=A \frac{\frac{3}{4} \times B}{2}=\frac{3 A B}{8}
$
MCQ 151 Mark
A magnetic field of $2 \times 10^{-2} T$ acts right angles to a coil of area $100 cm ^2$ with 50 turns. The average e.m.f. induced in the coil is $0.1 V$, when it is removed from the field in time ' $t$ '. The value of ' $t$ ' is
- A$2 \times 10^{-3} s$
- B$0.5 s$
- ✓$0.1 s$
- D$1 s$
Answer
View full question & answer→Correct option: C.
$0.1 s$
(c) : As, $\varepsilon=\frac{-d \phi}{d t}=-\frac{(0-N B A)}{t}$
$
t=\frac{N B A}{\varepsilon}=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{0.1} ; t=0.1 s
$
$
t=\frac{N B A}{\varepsilon}=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{0.1} ; t=0.1 s
$
MCQ 161 Mark
A horizontal wire of mass ' $m$ ', length ' $l$' and resistance ' $R$ ' is sliding on the vertical rails on which uniform magnetic field ' $B$ ' is directed perpendicular. The terminal speed of the wire as it falls under the force of gravity is ( $g=$ acceleration due to gravity)
- A$\frac{m g l}{B R}$
- B$\frac{B^2 l^2}{m g R}$
- C$\frac{m g R}{B l}$
- ✓$\frac{m g R}{B^2 l^2}$
Answer
View full question & answer→Correct option: D.
$\frac{m g R}{B^2 l^2}$
(d) : Magnetic force, $F=B I l$
$m g=B l \cdot \frac{E}{R}$. . . . . .(i)
Putting, emf, $E=B v l$ in eq (i),
So, $m g=\frac{B l}{R} \cdot B v l ; m g=\frac{B^2 l^2}{R} v \Rightarrow v=\frac{m g R}{B^2 l^2}$
$m g=B l \cdot \frac{E}{R}$. . . . . .(i)
Putting, emf, $E=B v l$ in eq (i),
So, $m g=\frac{B l}{R} \cdot B v l ; m g=\frac{B^2 l^2}{R} v \Rightarrow v=\frac{m g R}{B^2 l^2}$
MCQ 171 Mark
The alternating e.m.f. induced in the secondary coil of a transformer is mainly due to
- Avarying electric field
- ✓varying magnetic field
- Cthe iron core
- Dheat produced in the coil.
Answer
View full question & answer→Correct option: B.
varying magnetic field
(b) : When magnetic field changes, the magnetic flux also changes and then in a coil emf is induced.
MCQ 181 Mark
- ✓$2 B L \sin \left(\frac{\theta}{2}\right)(g L)^{1 / 2}$
- B$B L \sin \left(\frac{\theta}{2}\right)(g L)$
- C$2 B L \sin \left(\frac{\theta}{2}\right)(g L)^{3 / 2}$
- D$2 B L \sin \left(\frac{\theta}{2}\right)(g L)^2$
Answer
View full question & answer→Correct option: A.
$2 B L \sin \left(\frac{\theta}{2}\right)(g L)^{1 / 2}$
(a) : As, $h=L(1-\cos \theta)$. . . . . .(i)
At equilibrium, maximum velocity is given by, $v^2=2 g h=2 g L(1-\cos \theta)$
$v^2=2 g L\left(2 \sin ^2 \frac{\theta}{2}\right) ; v=2 \sqrt{g L} \sin \left(\frac{\theta}{2}\right)$
Maximum potential difference is,
$
\begin{aligned}
& V_{\max }=B \nu L \\
& =B \times 2 \sqrt{g L} \sin \left(\frac{\theta}{2}\right) \cdot L=2 B L \sin \left(\frac{\theta}{2}\right)(\sqrt{g L})
\end{aligned}
$
At equilibrium, maximum velocity is given by, $v^2=2 g h=2 g L(1-\cos \theta)$
$v^2=2 g L\left(2 \sin ^2 \frac{\theta}{2}\right) ; v=2 \sqrt{g L} \sin \left(\frac{\theta}{2}\right)$
Maximum potential difference is,
$
\begin{aligned}
& V_{\max }=B \nu L \\
& =B \times 2 \sqrt{g L} \sin \left(\frac{\theta}{2}\right) \cdot L=2 B L \sin \left(\frac{\theta}{2}\right)(\sqrt{g L})
\end{aligned}
$
MCQ 191 Mark
The self inductance ' $L$ ' of a solenoid of length ' $T$ ' and area of cross-section ' $A$ ' with a fixed number of turns ${ }^{\text {' } N}$, increases as
- ✓$I$ decreases and $A$ increases
- Bboth $l$ and $A$ decrease
- Cboth $l$ and $A$ increase
- D$l$ increases and $A$ decreases.
Answer
View full question & answer→Correct option: A.
$I$ decreases and $A$ increases
(a) : The self inductance ' $L$ ' of a solenoid of length ' $l$ ' and area of cross-section ' $A$ ' with $N$ number of turns is,
$L=\frac{\mu_0 N^2 A}{l}$. . . . . . .(i)
From equation (i), ' $L$ ' will be increases when area increases and ' $T$ decreases.
$L=\frac{\mu_0 N^2 A}{l}$. . . . . . .(i)
From equation (i), ' $L$ ' will be increases when area increases and ' $T$ decreases.
MCQ 201 Mark
Two coils $A$ and $B$ are placed in a circuit. When current in the coil $A$ changes by $0.8 A$ the magnetic flux in coil $B$ changes by $0.16 Wb$. The mutual inductance between the coils is
- ✓$0.2 H$
- B$20 H$
- C$0.5 H$
- D$2 H$
Answer
View full question & answer→Correct option: A.
$0.2 H$
(a) : As, $\phi=m i ; \quad M=\frac{\phi}{I}=\frac{0.16}{0.8}=0.2 H$
MCQ 211 Mark
If flux is given as $\phi=3 t^2+4 t+8$. Then find the induced emf at $t=2 s$
- ✓$16 V$
- B$12 V$
- C$8 V$
- D$4 V$
Answer
View full question & answer→Correct option: A.
$16 V$
(a) : $\phi=3 t^2+4 t+8$
$
e=\frac{d \phi}{d t}=6 t+4 ;\left.e\right|_{t=2 s}=16 V
$
$
e=\frac{d \phi}{d t}=6 t+4 ;\left.e\right|_{t=2 s}=16 V
$
MCQ 221 Mark
An infinitely long cylinder is kept parallel to a uniform magnetic field $B$ directed along positive $z$-axis. The direction of induced current on the surface of cylinder as seen from the $z$-axis will be
- Aclockwise of the +ve $z$-axis
- Banticlockwise of the $+ ve z$-axis
- ✓zero
- Dalong the magnetic field/Activate
Answer
View full question & answer→Correct option: C.
zero
(c) : Since the magnetic field is uniform, hence there is no change in the magnetic flux linked with cylindrical wire and hence no current will be induced on the surface of cylindrical wire.
MCQ 231 Mark
Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1>R_2$, the mutual inductance $M$ between them will be directly proportional to
- A$\frac{R_1}{R_2}$
- B$\frac{R_2}{R_1}$
- C$\frac{R_1^2}{R_2}$
- ✓$\frac{R_2^2}{R_1}$
Answer
View full question & answer→Correct option: D.
$\frac{R_2^2}{R_1}$
(d) : Let a current $I_1$ flows through the outer circular coil of radius $R_1$.
The magnetic field at the centre of the coil is $B_1=\frac{\mu_0 I_1}{2 R_1}$
As the inner coil of radius $R_2$ placed co-axially has smaller radius $\left(R_2<R_1\right)$, therefore $B_1$ may be taken to be constant over its cross-sectional area.
Hence, flux associated with the inner coil is
$\phi_2=B_1 \pi R_2^2=\frac{\mu_0 I_1}{2 R_1} \pi R_2^2$
As $M=\frac{\phi_2}{I_1}=\frac{\mu_0 \pi R_2^2}{2 R_1} \quad \therefore M \propto \frac{R_2^2}{R_1}$
The magnetic field at the centre of the coil is $B_1=\frac{\mu_0 I_1}{2 R_1}$
As the inner coil of radius $R_2$ placed co-axially has smaller radius $\left(R_2<R_1\right)$, therefore $B_1$ may be taken to be constant over its cross-sectional area.
Hence, flux associated with the inner coil is
$\phi_2=B_1 \pi R_2^2=\frac{\mu_0 I_1}{2 R_1} \pi R_2^2$
As $M=\frac{\phi_2}{I_1}=\frac{\mu_0 \pi R_2^2}{2 R_1} \quad \therefore M \propto \frac{R_2^2}{R_1}$
MCQ 241 Mark
Two coils have a mutual inductance of $0.01 H$. The current in the first coil changes according to equation $I=5 \sin 200 \pi t$. The maximum value of e.m.f. induced in the second coil is
- ✓$10 \pi$ volt
- B$0.1 \pi$ volt
- C$\pi$ volt
- D$0.01 \pi$ volt
Answer
View full question & answer→Correct option: A.
$10 \pi$ volt
(a) : The emf induced in the second coil is $e=-M \frac{d I}{d t}$. . . . . . .(i)
where $M$ is the mutual inductance and $\frac{d I}{d t}$ is the rate of change of current in the first coil.
Given, $I=5 \sin 200 \pi t$
$\therefore \frac{d I}{d t}=-5 \times 200 \pi \cos 200 \pi t=-1000 \pi \cos 200 \pi t$ $=1000 \pi \sin \left(200 \pi t-\frac{\pi}{2}\right)$
From equation (i),
$e=0.01 \times 1000 \pi \sin (200 \pi t-\pi / 2)=10 \pi \sin \left(200 \pi t-\frac{\pi}{z}\right)$
For maximum value of the induced emf,
$
\sin (200 \pi t-\pi / 2)=1 \quad \therefore \quad e_{\max }=10 \pi V
$
where $M$ is the mutual inductance and $\frac{d I}{d t}$ is the rate of change of current in the first coil.
Given, $I=5 \sin 200 \pi t$
$\therefore \frac{d I}{d t}=-5 \times 200 \pi \cos 200 \pi t=-1000 \pi \cos 200 \pi t$ $=1000 \pi \sin \left(200 \pi t-\frac{\pi}{2}\right)$
From equation (i),
$e=0.01 \times 1000 \pi \sin (200 \pi t-\pi / 2)=10 \pi \sin \left(200 \pi t-\frac{\pi}{z}\right)$
For maximum value of the induced emf,
$
\sin (200 \pi t-\pi / 2)=1 \quad \therefore \quad e_{\max }=10 \pi V
$
MCQ 251 Mark
An ideal transformer converts $220 V$ a.c. to $3.3 kV$ a.c. to transmit a power of $4.4 kW$. If primary coil has 600 turns, then alternating current in secondary coil is
- A$\frac{1}{3} A$
- ✓$\frac{4}{3} A$
- C$\frac{5}{3} A$
- D$\frac{7}{3} A$
Answer
View full question & answer→Correct option: B.
$\frac{4}{3} A$
(b) : Given, $V_{\text {in }}=220 V , V_{\text {out }}=3.3 \times 10^3 V$
Power $=4.4 kW ; \quad N_P=600 ;$ Power input is, $P=V_{\text {im }} \times R_{\text {in }}$
$
R_{in}=\frac{4.4 \times 1000}{220}=\frac{44 \times 10}{22}=20 A
$
Since for an ideal transformer
$
\frac{V_{\text {out }}}{V_{\text {in }}}=\frac{R_p}{R_s} \Rightarrow R_s=R_p \times \frac{V_{\text {out }}}{V_{\text {in }}}=\frac{20 \times 220}{3.3 \times 1000}=\frac{44}{33}=\frac{4}{3} A
$
Power $=4.4 kW ; \quad N_P=600 ;$ Power input is, $P=V_{\text {im }} \times R_{\text {in }}$
$
R_{in}=\frac{4.4 \times 1000}{220}=\frac{44 \times 10}{22}=20 A
$
Since for an ideal transformer
$
\frac{V_{\text {out }}}{V_{\text {in }}}=\frac{R_p}{R_s} \Rightarrow R_s=R_p \times \frac{V_{\text {out }}}{V_{\text {in }}}=\frac{20 \times 220}{3.3 \times 1000}=\frac{44}{33}=\frac{4}{3} A
$
MCQ 261 Mark
The magnetic flux near the axis and inside the air core solenoid of length $60 cm$ carrying current $I$ is $1.57 \times 10^{-6} Wb$. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $\mu_0=4 \pi \times 10^{-7} SI$ unit)
- A$0.25 A$
- B$0.50 A$
- ✓$0.75 A$
- D$1 A$
Answer
View full question & answer→Correct option: C.
$0.75 A$
(c) : Given, $l=60 cm , \phi=1.57 \times 10^{-6} Wb$
Magnetic induction inside the solenoid $B=\frac{\mu_0 n I}{l}$
$
\begin{aligned}
& \text { Magnetic flux } \phi=B A=\frac{\mu_0 n I A}{l} \\
& \text { Magnetic moment }=n I A=\frac{\phi l}{\mu_0} \\
& =\frac{1.57 \times 10^{-6} \times 60 \times 10^{-2}}{4 \times 3.14 \times 10^{-7}}=0.75 A
\end{aligned}
$
Magnetic induction inside the solenoid $B=\frac{\mu_0 n I}{l}$
$
\begin{aligned}
& \text { Magnetic flux } \phi=B A=\frac{\mu_0 n I A}{l} \\
& \text { Magnetic moment }=n I A=\frac{\phi l}{\mu_0} \\
& =\frac{1.57 \times 10^{-6} \times 60 \times 10^{-2}}{4 \times 3.14 \times 10^{-7}}=0.75 A
\end{aligned}
$
MCQ 271 Mark
Two coils $P$ and $Q$ are kept near each other. When no current flows through coil $P$ and current increases in coil $Q$ at the rate $10 A / s$, the e.m.f. in coil $P$ is $15 mV$. When coil $Q$ carries no current and current of 1.8 A flows through coil $P$, the magnetic flux linked with the coil $Q$ is
- A$1.4 mWb$
- B$2.2 mWb$
- ✓$2.7 mWb$
- D$2.9 mWb$
Answer
View full question & answer→Correct option: C.
$2.7 mWb$
(c) : The emf produced in the coil is given
$
\varepsilon=M \frac{d I}{d t}
$
Given : $\varepsilon=15 mV , \frac{d I}{d t}=10 A / s , l=1.8 A$
$
15 \times 10^{-3}=M \times 10 ; M=\frac{15 \times 10^{-3}}{10}=15 \times 10^{-4} H
$
The magnetic flux $=$ MI
$
=15 \times 10^{-4} \times 1.8=27 \times 10^{-4}=2.7 \times 10^{-3}=2.7 mWb
$
$
\varepsilon=M \frac{d I}{d t}
$
Given : $\varepsilon=15 mV , \frac{d I}{d t}=10 A / s , l=1.8 A$
$
15 \times 10^{-3}=M \times 10 ; M=\frac{15 \times 10^{-3}}{10}=15 \times 10^{-4} H
$
The magnetic flux $=$ MI
$
=15 \times 10^{-4} \times 1.8=27 \times 10^{-4}=2.7 \times 10^{-3}=2.7 mWb
$
MCQ 281 Mark
Magnetic flux passing through a coil is initially $4 \times 10^{-4} \ Wb$. It reduces to $10 \%$ of its original value in $t$ second. If the $e.m.f.$ induced is $0.72 \ mV$ then $t$ in second is
- A$0.3$
- B$0.4$
- ✓$0.5$
- D$0.6$
Answer
View full question & answer→Correct option: C.
$0.5$
$e=-\frac{\left(\phi_2-\phi_1\right)}{t}$
$\therefore t=\frac{\left(\phi_1-\phi_2\right)}{e}$
$=\frac{4 \times 10^{-4}-0.4 \times 10^{-4}}{0.72 \times 10^{-3}}$
$=\frac{3.6 \times 10^{-4}}{0.72 \times 10^{-3}}$
$=0.5 s$
$\therefore t=\frac{\left(\phi_1-\phi_2\right)}{e}$
$=\frac{4 \times 10^{-4}-0.4 \times 10^{-4}}{0.72 \times 10^{-3}}$
$=\frac{3.6 \times 10^{-4}}{0.72 \times 10^{-3}}$
$=0.5 s$
MCQ 291 Mark
Alternating current of peak value $\left(\frac{2}{\pi}\right)$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=50 Hz$ )
- A$100 V$
- ✓$200 V$
- C$300 V$
- D$400 V$
Answer
View full question & answer→Correct option: B.
$200 V$
(b): Let $I_p=\frac{2}{\pi} \sin \omega t A$
Emf induced in the secondary coil
$|e|=M \frac{d I_p}{d t}=M \frac{d}{d t}\left(\frac{2}{\pi} \sin \omega t\right)=\frac{2 M \omega}{\pi} \cos \omega t$
$\therefore \quad$ Peak emf induced in secondary coil is
$
|e|_{\max }=\frac{2 M \omega}{\pi}=\frac{2 \times 1 \times 2 \pi \times 50}{\pi}=200 V
$
Emf induced in the secondary coil
$|e|=M \frac{d I_p}{d t}=M \frac{d}{d t}\left(\frac{2}{\pi} \sin \omega t\right)=\frac{2 M \omega}{\pi} \cos \omega t$
$\therefore \quad$ Peak emf induced in secondary coil is
$
|e|_{\max }=\frac{2 M \omega}{\pi}=\frac{2 \times 1 \times 2 \pi \times 50}{\pi}=200 V
$
MCQ 301 Mark
Two coils $A$ and $B$ have mutual inductance $2 \times 10^{-2}$ henry. If the current in the primary is $i=5 \sin (10 \pi t)$ then the maximum value of $emf$ induced in coil $B$ is
- ✓$\pi$ volt
- B$\frac{\pi}{2}$ volt
- C$\frac{\pi}{3}$ volt
- D$\frac{\pi}{4}$ volt
Answer
View full question & answer→Correct option: A.
$\pi$ volt
(a) : Here, $M=2 \times 10^{-2} H , i=5 \sin (10 \pi t)$
Maximum emf induced, $\varepsilon_0=$ ?
emf induced in the second coil (B),
$
\begin{aligned}
& \varepsilon=M \frac{d i}{d t}=M \frac{d}{d t}(5 \sin (10 \pi t)) \\
& =M \times 5 \times 10 \pi \times \cos (10 \pi t) \\
& =2 \times 10^{-2} \times 5 \times 10 \pi \times \cos (10 \pi t)=\varepsilon_0 \cos (10 \pi t) \\
& \therefore \quad \varepsilon_0=2 \times 10^{-2} \times 5 \times 10 \pi=\pi \text { volt }
\end{aligned}
$
Maximum emf induced, $\varepsilon_0=$ ?
emf induced in the second coil (B),
$
\begin{aligned}
& \varepsilon=M \frac{d i}{d t}=M \frac{d}{d t}(5 \sin (10 \pi t)) \\
& =M \times 5 \times 10 \pi \times \cos (10 \pi t) \\
& =2 \times 10^{-2} \times 5 \times 10 \pi \times \cos (10 \pi t)=\varepsilon_0 \cos (10 \pi t) \\
& \therefore \quad \varepsilon_0=2 \times 10^{-2} \times 5 \times 10 \pi=\pi \text { volt }
\end{aligned}
$
MCQ 311 Mark
Same current is flowing in two a.c. circuits. First contains only inductance and second contains only capacitance. If frequency of a.c. is increased for both, the current will
- Aincrease in first circuit and decrease in second
- Bincrease in both circuits
- Cdecrease in both circuits
- ✓decrease in first circuit and increase in second
Answer
View full question & answer→Correct option: D.
decrease in first circuit and increase in second
(d) : Capacitive reactance
$
X_C=\frac{1}{\omega C}=\frac{1}{2 \pi \nu C}
$
Inductive reactance, $X_L=\omega L=2 \pi \omega L$
When frequency of a.c. source is increased, inductive reactance increases while capacitive reactance decreases. Also, Current $\propto \frac{1}{\text { Impedance }}$
So, current in inductive circuit decreases and in capacitive circuit increases on increasing the frequency of a.c. source.
$
X_C=\frac{1}{\omega C}=\frac{1}{2 \pi \nu C}
$
Inductive reactance, $X_L=\omega L=2 \pi \omega L$
When frequency of a.c. source is increased, inductive reactance increases while capacitive reactance decreases. Also, Current $\propto \frac{1}{\text { Impedance }}$
So, current in inductive circuit decreases and in capacitive circuit increases on increasing the frequency of a.c. source.
MCQ 321 Mark
If $N$ is the number of turns in a circular coil then the value of self inductance varies as
- A$N^0$
- B$N$
- ✓$N ^2$
- D$N ^{-2}$
Answer
View full question & answer→Correct option: C.
$N ^2$
(c) : Self inductance of a circular coil is
$
L=\frac{1}{2} \mu_0 N^2 \pi R
$
where, $N=$ Number of turns in the coil $R=$ Radius of the coil $\therefore L \propto N^2$
$
L=\frac{1}{2} \mu_0 N^2 \pi R
$
where, $N=$ Number of turns in the coil $R=$ Radius of the coil $\therefore L \propto N^2$