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CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 4 - MCQExams.com
CBSE
Class 12 Medical Physics
Electromagnetic Induction
Quiz 4
The coefficient of mutual induction between two coils is 4 H. If the current in the primary reduces from 5A to zero in 10$$^{-3}$$ second then the induced emf in the secondary coil will be
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$$10^4$$ V
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$$25 \times 10^3$$ V
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$$2 \times 10^4 $$ V
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$$15 \times 10^3$$ V
Explanation
emf = $$ - M \dfrac{dI}{dt}$$
$$ = -4 \dfrac{0-5}{10^{-3} }= 2 \times 10^4 V $$
A magnet is brought near a coil in two ways (i) rapidly (ii) slowly. The induced charge will be
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0%
More in case (i)
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More in case (ii)
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Equal in both the cases
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More or less according to the radius of the coil
The SI unit of inductance, the henry, can be written as :
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weber / ampere
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volt second / ampere
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joule / ampere$$^2$$
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ohm second
Explanation
$$ L\dfrac{dI}{dt} = emf = \dfrac{d\phi}{dt}$$
a) $$ [L] = weber / ampere $$
b) $$ [L] = volt.second / ampere $$
c) $$ Joule = volt \times ampere \times second $$
$$ \Rightarrow volt \times second = Joule / ampere$$
$$ \Rightarrow [L] = joule/ ampere^2$$
d) $$ V = IR \Rightarrow volt = ampere.ohm$$
$$\Rightarrow [L] = ohm.second$$.
A straight copper wire is moved in a uniform magnetic field such that it cuts the magnetic lines of force. Then
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emf will not be induced
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emf will be induced
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sometimes emf will be induced and sometimes not
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nothing can be predicted
Explanation
As the copper wire cuts the magnetic lines of force, the free charges in the wire will experience a force and will get displaced. This displacement of charges in the wire creates a potential difference and hence emf is induced.
The coefficient of mutual induction between two coils is $$1.25 H.$$ If the rate of fall of current in the primary is $$80$$ As$$^{-1}$$, then the induced emf in the secondary coil will be
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$$100 V$$
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$$64 V$$
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$$12.5 V$$
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$$0.016 V$$
Explanation
emf = $$ - M \dfrac{dI}{dt}$$
$$ = -1.25 \times -80= 100 V $$
A coil of insulating wire is connected to battery. If it is moved towards a galvanometer then its point gets deflected because
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the coil behaves like a magnet
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induced current is produced in the coil
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the number of turns in the galvanometer coil remains constant
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none of the above
Explanation
Upon connecting to the battery , a current starts to flow in the insulated wire. This current induces a current in the galvanometer coil due to mutual inductance because of the flux linkages. This induced current causes a deflection in the galvanometer .
The rate of change of magnetic flux density through a circular coil of area 10 m and number of turns 100 is 10$$^3$$ Wb/m$$^2$$/s. The value of induced emf will be
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10$$^{-2}$$ V
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10$$^{-3}$$ V
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10 V
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10$$^{6}$$ V
Explanation
$$emf =$$ rate of change of flux
$$=$$ area $$\times $$ number of turns $$\times $$ rate of change of flux density
$$= 10 \times 100 \times 1000 \\ =10^6 V $$
The distance between the ends of the wings of an airplane is $$50 m$$. It is flying in a horizontal plane at a speed of $$360 Km/hour$$. The vertical component of earth's magnetic field at that place is $$2.0 \times 10^{-4} Wb/m^2$$, then the potential difference induced between the ends of the wings will be
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0%
$$0.1 volt$$
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$$1.0 volt$$
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$$0.2 volt$$
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$$0.01 volt$$
Explanation
$$emf = B l v $$
$$= 2 \times 10^{-4} \times 50 \times 360 \times \dfrac{ 1000}{3600} = 1 V $$
If the turns ratio of a transformer is 2 and the impedance of primary coil is 250 W then the impedance of secondary coil will be
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1000 $$\Omega$$
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500 $$\Omega$$
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250 $$\Omega$$
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125 $$\Omega$$
Explanation
As shown in the image $$ L \propto N^2 $$
and transformer has coils which cause impedance and impedance of a inductor is $$2 \pi fL$$
so impedance of each coil $$ \propto N^2 $$
so impedance of secondary coil $$={ Z }_{ p }{ (\dfrac { { N }_{ S } }{ { N }_{ P } } })^{ 2 }\\=250*2^2 = 1000 \Omega$$
The coefficient of self induction of a coil is given by
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$$\displaystyle L = \left ( -\frac{dI}{dt} \right )$$
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$$\displaystyle L = - \frac{edI}{dt}$$
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$$\displaystyle L = \frac{dI}{edt}$$
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$$\displaystyle L = \frac{dI}{dt}e^2$$
Explanation
Inductance is the property of a conductor by which a change in current flowing through it induces (creates) a voltage (electromotive force) in both the conductor itself (self-inductance) and in any nearby conductors (mutual inductance). By Lenz's law the induced voltage opposed the the change in current. Hence inductance is defined as
$$L = - \dfrac{dI}{dt} $$
The length of side of a square coil is 50 cm and number of turns in it isIf it is placed at right angles to a magnetic field which is changing at the rate of 4 Tesla/s, then induced emf in the coil will be :
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0.1 V
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1.0 V
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10 V
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100 V
Explanation
emf =|e|= $$\dfrac{d\phi}{dt}$$=NA$$\dfrac{dB}{dt}$$
rate of change of flux = $$N \times area \times $$ rate of change of magnetic field = $$100 \times 0.5^2 \times 4 = 100 V $$
If a spark is produced on removing the load from an AC circuit then the element connected in the circuit is
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high resistance
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high capacitance
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high inductance
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high impedance
Explanation
On removal of load from the circuit, the circuit suddenly becomes an open circuit.
Thus $$ \dfrac{di}{dt} \rightarrow \infty $$
For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the inductor is $$ L\dfrac{di}{dt} $$
Therefore, the circuit has high inductance.
When a coil of cross-sectional area $$A$$ and number of turns $$N$$ is rotated in a uniform magnetic field $$B$$ with angular velocity $$\omega$$, then the maximum emf induced in the coil will be
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$$BNA$$
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$$\displaystyle \frac{B A \omega}{N}$$
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$$BNA$$ $$\omega$$
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$$zero$$
Explanation
$$emf =$$ rate of change of flux
$$=BAN\omega$$
The magnetic fields through two identical rings made of copper and wood are changing at the same rate. The induced electric field in copper ring will be :
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more than that in the wooden ring
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less than that in the wooden ring
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finite and that in the wooden ring will be zero
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same as that in the wooden ring
Explanation
emf induced = rate of change of flux = $$ \frac{ d (\vec A . \vec B) }{dt} $$. As the two rings are identical the area of both the rings are same. Hence the flux and therefore the rate of change of flux in both is the same. Therefore emf induced in both is same.
Electric field = $$ \frac{emf}{ length} $$. As both the rings are identical, their length is same.
And hence the electric field in both is the same.
The resistance coils in a resistance box are made of double folded wire so that their
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self induction effect in nullified
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self inductance is maximum
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induced emf is maximum
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None of these
Explanation
Presence of inductance increases the impedance more than the actual resistance value. Therefore to reduce the chance of occurrence of inductance in the resistor coils, they are double folded, so that current in opposite direction cancel the induced flux linkages if any and hence inductance is nullified.
A conducting rod of length L is falling with velocity V in a uniform horizontal magnetic field B normal to the rod. The induced emf between the ends the rod will be :
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2 BV$$l$$
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zero
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B$$l$$V
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$$\displaystyle \frac{BVl}{2}$$
Explanation
Force on charge q due to the motion of rod in the field= $$ F = q V B $$
This force on the charge is attributed to the induced electric field $$E$$
Therefore,
$$ Eq = F = q V B $$
$$ E = V B $$
therefore, an electric field $$ E = V B $$ is said to be induced in the rod due to the motion in the magnetic field.
Potential difference due to the field = emf induced =$$ E \times length = E l = VBl volts $$
A millivoltmeter is connected in parallel to an axle of the train running with a speed of 180 km/hour. If the vertical component of earth's magnetic field is $$0.2 \times 10^{-4} Wb/m^2$$ and the distance between the rails is 1m, then the reading of voltmeter will be :
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10$$^{-2}$$ volt
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10$$^{-4}$$ volt
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10$$^{-3}$$ volt
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1 volt
Explanation
Reading of voltmeter= emf induced = $$ B l v = 0.2 \times 10^{-4} \times 1 \times 180 \times \dfrac{1000}{3600}=1 mV $$
What will happen inside the conductor, If a permanent magnet is sitting next to a conductor:
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there will be no induced voltage in the conductor
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the induced voltage in the conductor will increase
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the induced voltage in the conductor will decrease
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the magnetic field will induce a voltage in the conduct
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the induced voltage in the conductor will increase momentarily, but then decrease rapidly
Explanation
Voltage is induced in an conductor when the electrons in the conductor experience a magnetic force due to external magnetic field and thus collect at one side of the conductor. However a stationary conductor would have no electron experience that force. Hence there would be no induced voltage.
The coefficients of self induction of two coils are $$L_1 = $$ 8mH and $$L_2=$$ 2mH respectively. The current rises in the two coils at the same rate. The power given to the two coils at any instant is same. The ratio of induced emf's in the coils will be :
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$$\displaystyle \frac{V_1}{V_2} = 4$$
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$$\displaystyle \frac{V_1}{V_2} = \frac{1}{4}$$
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$$\displaystyle \frac{V_1}{V_2} = \frac{1}{2}$$
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$$\displaystyle \frac{V_1}{V_2} = \frac{1}{3}$$
Explanation
Power to 1st coil = $$ V_1 i_1 = L_1 \dfrac{di_1}{dt} i_i $$
Power to 2nd coil = $$ V_2 i_2 = L_2 \dfrac{di_2}{dt} i_2 $$
As the powers are same,
$$ L_1 \dfrac{di_1}{dt} i_i = L_2 \dfrac{di_2}{dt} i_2 $$
rate of change of current is given to be same,
$$ L_1 i_i = L_2 i_2 $$
or, $$ \dfrac {i_1}{i_2} = \dfrac{ L_2}{L_1} =\dfrac{ 1}{4} $$
So, $$ \dfrac{ V_1}{V_2} = \dfrac{ L_1 \dfrac{di_1}{dt} } { L_2 \dfrac{di_2}{dt} } = \dfrac{ L_1}{L_2} = 4 $$
The number of turns in a coil of wire of fixed radius is 600 and its self inductance is 108 mH. The self inductance of a coil of 500 turns will be :
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74 mH
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75 mH
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76 mH
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77 mH
Explanation
$$L = \dfrac{\mu n^2 A}{L}$$.
$$L \propto n^2$$
Thus, for 500 turns, self inductance would be $$108\times (500)^2/(600)^2=108\times 25/36=75mH$$
The magnetic flux linked with a coil is $$\phi \leq 8t^2 +3t + 5$$ Weber. The induced emf in fourth second will be
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$$16 V$$
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$$139 V$$
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$$67 V$$
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$$145 V$$
Explanation
emf(t) = $$ \dfrac{ d \phi}{dt} = 16t+3 $$
emf(4)= $$16 \times 4 +3 = 67 V $$
An artificial with a metal surface is moving about the earth in a circular orbit. An emf (neglecting declination) will be induced in the satellite, if
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the plane of the orbit coincides with the equatorial plane.
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the plane of the orbit is inclined to the equatorial plane.
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the plane of the orbit coincides with the equatorial plane and the speed of the satellite is less than 8 kmsec$$^{-1}$$.
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no induced emf produced whatever may be the plane of the orbit.
Explanation
If the latitude of earth remains the same in the orbit, then the flux through the loop made by the artificial satellite remains the same and hence no emf is induced.
magnetic flux density of the earth changes with the latitude. Therefore if the plane of the orbit of the sattelite is inclined at an angel to the equatorial plane, the flux density changes along the orbit and the flux changes and an emf would be induced.
The value of mutual inductance can be increased by
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decreasing N
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increasing N
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winding the coil on wooden frame
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winding the coil on china clay
Explanation
$$M = \mu N_1 N_2 \dfrac{ Area_{12}}{length_{12}} $$
$$ Area_{12} $$ - area in common to both the coils where the flux links both of them together.
$$length_{12}$$ - length in common to both the coils where the flux links both of them together.
Therefore to increase the mutual inductance, the number of turns can be increased.
A cylindrical bar magnet is lying along the axis of a circular coil. If the magnet is rotated about the axis of the coil then
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emf will be induced in the coil
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only induced current will be generate in the coil
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no current will be induced in the coil
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both emf and current will be induced in the coil
Explanation
As cylindrical bar magnet is lying along the axis of a circular coil, if the magnet bar is rotated about the axis of the coil, due to symmetry of the arrangement, there will be no change in flux. As there would be no change in flux, there will be no induced emf , nor any induced current.
The value of coefficient of mutual induction for the arrangement of two coils shown in the figure will be :
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Zero
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Maximum
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Negative
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Positive
Explanation
The mutual inductance between two coils depends upon the manner, in which, two coils are placed relative to each other. In the given figure, the magnetic flux linked with a coil due to current in another coil, seems to be zero, therefore coefficient of mutual inductance will be zero.
The magnetic flux in a coil of 100 turns increases by $$12 \times 10^3$$ Maxwell in 0.2 second due to the motion of a magnet. The emf induced in the coil will be
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6 V
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0.6 V
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0.06 V
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60 V
Explanation
$$emf =$$ rate of change of flux $$\times $$ number of turns $$= \dfrac{12 \times 10^3 \times 10^{-8}}{0.2} \times 100 = 0.06 V $$
1 Maxwell = $$ 10^{-8} $$ weber
The two rails of a railways track, insulated from each other and the ground, are connected to a milli voltmeter. What is the reading of the milli voltmeter when a train travels at a speed of 20 ms$$^{-1}$$ along the track, given that the vertical component of the earth's magnetic field is 0.2 $$\times$$10$$^{-4}$$ Wbm$$^{-2}$$ and the rails are separated by 1 m?
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4 mV
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0.4 mV
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80 mV
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10 mV
Explanation
emf = $$ B l v $$
$$=0.2 \times 10 ^{-4} \times 1 \times 20 $$
$$= 0.4 m V $$
Flux $$\varphi$$ (in water) in a closed circuit of resistance 10 $$\Omega$$ varies with time t (in sec) according to the equation $$\varphi = 6t^2 - 5t + 1$$. What is the magnitude of the induced current at $$t = 0.25$$s?
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1.2 A
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0.8 A
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0.6 A
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0.2 A
Explanation
$$emf = -\dfrac{d \phi}{dt}= -12t + 5 $$
current = $$- \dfrac{emf}{10} = -1.2 t + 0.5 $$
current at $$t= 0.25 s$$,
$$=-1.2 \times 0.25 + 0.5 =0.2 A$$
The coefficient of mutual inductance of the two coils is $$5H$$. The current through the primary coil is reduced to zero value from $$3 A$$ in $$1 $$millisecond. The induced emf in the secondary coils is
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$$zero$$
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$$1.67 KV$$
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$$15 KV$$
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$$600 V$$
Explanation
Flux linking the secondary coil due to current in primary $$= M i_1$$
induced emf in the secondary coils $$=$$ rate of change of flux
$$ =M \dfrac{\triangle i_1}{\triangle t} $$
$$= 5 \times \dfrac{ 3-0}{0.001}$$
$$=15 k V$$
A conducting rod of length L is falling with velocity v perpendicular to a uniform horizontal magnetic field B. The potential difference between its two ends will be
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2BLv
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BLv
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$$\frac{1}{2}$$BLv
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(BLv)$$^2$$
Explanation
Force on a charge q in the rod = $$ q v B $$
electric field inside the rod due to displacements of charges due to force by relative motion in magnetic field $$= \dfrac{emf}{l} $$
As the rod moves in constant velocity, net force of constituent charge q in the rod $$=0$$
Therefore,
$$ q\dfrac{emf}{l} = q v B$$
$$ \Rightarrow emf = B l v$$
The area of the coil must be._____
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$$1.8\ m^2$$
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$$18\ m^2$$
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$$8\ m^2$$
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$$none\ of\ these$$
Explanation
Flux through a circular coil $$\phi=NBA\cos \omega t$$
Voltage required $$\displaystyle \varepsilon = \dfrac{-d \phi}{dt} $$
$$\Rightarrow 9 = NBA \omega sin \omega t$$
$$\displaystyle 9 = \dfrac{8 \times 10^{-5} \times A \times 30 \times 2\pi \times 2000}{60}$$
$$\displaystyle A = \dfrac{9 \times 10^5}{50 \times 10^3} = 18m^2$$
A car moves up a plane road. The induced emf in the axle connecting the two wheels is maximum when it moves
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At the poles
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At equator
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Remains stationary
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No emf is induced at all
Explanation
Induced emf in the axle $$ =B l v$$
$$v -$$ velocity of car
$$l -$$ length of car
$$B -$$ component of magnetic field perpendicular to both l and v .
That is B is the vertical component of magnetic field.
Vertical component of magnetic field is maximum at the poles.
Therefore emf induced in the axle will be maximum at the poles.
A uniform magnetic field exists exists in the region given by $$\overrightarrow B = \left( {3\widehat i + 4\widehat j + 5\widehat k} \right)T$$. A rod of length 1 m placed along y-axis is moved along x-axis with constant velocity 1 m/s, then induced emf in the rod is.
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Zero
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5 V
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2 V
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4 V
A rectangle coil ABCD is rotated anticlockwise with a uniform angular velocity about an axis as shown in the figure. The axis of rotation of the coil as well as the magnetic field B are horizontal. The induced emf in the coil would be minimum when
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the plane of coil is horizontal.
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the plane of coil is vertical.
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the plane of coil makes an angle of 45$$^o$$ with the direction of the magnetic field.
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the plane of coil makes an angle of 30$$^o$$ with the field.
Explanation
Let at time $$t=0$$, the coil ABCD is vertical as shown in figure and at time $$t$$, plane of coil makes an angle $$\theta$$ with the vertical, then
$$\theta=\omega t$$, ($$\omega=$$ uniform angular velocity),
in this position , magnetic flux linked with coil will be,
$$\phi=BA\cos \theta$$, (
where, $$A=$$ area of coil,
)
or $$\phi=BA \cos \omega t$$,
now, differentiating this equation w.r.t. time, we get
$$\dfrac{d\phi}{dt}=\frac{d}{dt}(BA \cos\omega t)$$,
or $$\dfrac{d\phi}{dt}=-BA\omega\sin\omega t$$,
if $e$$ is the emf induced in coil then by Faraday's law,
$$e=-\dfrac{d\phi}{dt}=BA\omega\sin\omega t$$,
now if, $$\sin\omega t=1$$ (maximum),
then $$e=BA\omega =e_{0}$$ (maximum),
therefore $$e=e_{0}\sin\omega t$$
it is clear from equation that $$e$$ will be minimum when
$$\omega t=0$$,
or $$\theta=0$$, the plane of coil is vertical.
A small circular coil of radius 1 cm and number of turns 100 is placed inside a long solenoid of radius 5 cm and number of turns 8 per cm. The axis of the coil is parallel to the solenoid axis. Then, the coefficient of mutual inductance of the two coils is (in milli Henry)
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0.032
0%
0.064
0%
0.016
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zero
Explanation
Field inside solenoid = $$B= \mu N i $$
N - number of turns per length in solenoid $$= 800/m$$
Flux linking the smaller coil = $$\lambda= 100 \times \pi r^2 B = 100 \times \pi r^2 \mu N i $$
mutual inductance = $$ M=\dfrac{\lambda}{i} = 100 \times \pi r^2 \mu N = 3.15 m H$$
An athlete is running at a speed of 30 kmh$$^{-1}$$ towards east, holding a 3 m metallic rod horizontally. The horizontal component of the earth, magnetic field in this region is $$3 \times 10^{-4} $$ Tesla and the angle of dip is 30$$^o$$. Then, the emf induced across the ends of the rod is
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7.5 mV
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4.3 mV
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zero
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13 mV
Explanation
Component of field in the vertical direction = $$ B_|=3 \times 10^{-4} tan( 30)$$
vertical component of field is perpendicular to both velocity and length of rod
emf = $$ B_| l v = 4.3 mV$$
Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be
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maximum in situation (A)
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maximum in situation (B)
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maximum in situation (C)
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the same in all situations
Explanation
The mutual inductance $$M$$
M
is given by, $$\phi=MI$$
, where $$\phi$$
ϕ
is the flux through a coil due to the current $$i$$
i
in another coil. The flux through an area $$\vec{A}$$
A
due to a magnetic field $$\vec{B}$$
is given by y $$\phi=\vec{B}.\vec{A}$$
⃗
ϕ=B→⋅A→
. The $$\vec{B}$$
and $$\vec{A}$$
A→
are parallel in configuration (A) but perpendicular in configuration (B) and configuration (C). Hence, the flux and mutual inductance are maximum in configuration (A).
Coils in the resistance boxes are made from doubled up insulated wires
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to cancel the effect of self induction
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to nullify the heating effect
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to nullify the Peltier effect
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to reduce effective length of the wire
Explanation
The wire is doubled back on itself. As a result ,there are equal and opposite currents in each section of the coil. Therefore ,the coil has no net magnetic field and no net induced e.m.f.
Thus, it is done to minimize the inductance of coils.
Complete the following sentence:
The current is induced in a closed circuit only if there is _________.
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change in number of magnetic field lines linked with the circuit.
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no change in number of magnetic field lines linked with the circuit.
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change in number of gravitational field lines linked with the circuit.
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no change in number of gravitational field lines linked with the circuit.
Explanation
The current is induced in a closed circuit only if there is a change in number of magnetic field lines in the circuit due to electromagnetic induction according to Faraday's law.
In an a.c. generator the speed at which the coil rotates is doubled. How would this affect
the frequency of output voltage ?
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frequency is doubled
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frequency is halved
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frequency remains same
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cant say
Explanation
AC generators are made up of one or more magnetic pole pairs corresponding to their purposes.
The frequency of an AC generator is dependent on the number of pairs of poles and the speed of the coil rotation. The frequency of the induced AC current depends on the revolution of the coil since, rate of revolution is Revolutions Per Minute (RPM) and frequency is expressed as cycles per second, the frequency of the induced AC current will be the rate of revolution of the coil divided by 60 seconds.
That is, $$F=RPM\times number \ of\ pole\ pairs\div 60$$
When the Revolutions per minute (RPM) is doubled, then, obviously the frequency of the ac current will also be doubled as the RPM is multiplied with the number of pole pairs.
A magnet is moved towards a coil (i) quickly (ii) slowly, then the induced e.m.f. is
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larger in case (i)
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smaller in case (ii)
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equal to both the cases
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larger or smaller depending upon the radius of the coil
Explanation
According to Faraday's Law: $$E = \cfrac{d \phi}{dt}$$
If the magnet moves quickly, the rate of change of flux is also higher, hence the EMF induced is larger in case (i).
State whether given statement is True or False
A device which receives and then transmits electromagnetic signal in an artificial satellite is called transponder.
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True
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False
Explanation
Answer is A.
A transponder is an electronic device used to wirelessly receive and transmit electrical signals. Fittingly, its name is equally derived from the words transmitter and responder. This device functions by receiving a signal, called an interrogator because it is effectively asking for information, then automatically conveying a radio wave at a predetermined frequency. In order to broadcast a signal on a different frequency than the one received, a frequency converter is built in. By receiving and transmitting on different frequencies, the two signals can be detected simultaneously.
Hence, the statement is true.
A copper ring is suspended by a thread in a vertical plane. If one end of a magnet is brought horizontally towards the ring in plane of ring as shown, the ring will
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move towards the magnet.
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not change its position.
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move away from the magnet.
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first move towards and then move away from the magnet
Explanation
Answer is B.
The magnetic flux is given by $$\phi= BA cos \theta$$ where $$\theta$$ is angle between magnetic field and area of ring.
in this case the area vector is out of the plane, hence $$\theta=90^\circ{}$$,
$$\Rightarrow \phi =0$$
The ring will not move from the magnet, as there is no change in flux through ring and current in it will be zero. Ring will not behave as a magnet.
The factors on which the magnitude of induced e.m.f. depend on
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change in magnetic flux
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the time in which the magnetic flux changes
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resistance of the coil
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both A & B
Explanation
The Faraday's laws of electromagnetic induction says that the E.M.F. induced in a coil 'e' = -(rate of change of magnetic flux linkage)
where,
the Flux linkage =$$number \ of \ turns'N'\times magnetic \ field'B'\times area'A'\times cos\theta$$
where,
theta is angle between magnetic field B and area A.
Theta at any instant 't'=$$(angular \ velocity \ w)\times (time \ instant't')$$. That is, Theta = $$w\times t$$.
E.M.F. induced in a coil 'e'=$$N\times B\times A\times w\times sinw\times t$$.
The factors involved in the induced emf of a coil are:
The induced e.m.f. is directly proportional to N, the total number of turns in the coil.
The induced e.m.f. is directly proportional to A, the area of cross-section of the coil.
The induced e.m.f. is directly proportional to B, the strength of the magnetic field in which the coil is rotating.
The induced e.m.f. is directly proportional to 'w', the angular velocity of coil.
The induced e.m.f. also varies with time and depends on instant 't'.
The induced e.m.f. is maximum when plane of coil is parallel to magnetic field B and e.m.f. is zero when plane of coil is perpendicular to magnetic field B.
The magnitude of induced current in a closed coil increases with the increase in the ................. of magnetic lines of force.
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strength
0%
alternating
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magnetic field
0%
less
Explanation
The magnitude of induced current in a closed coil increases with the increase in the
strength
of magnetic lines of force.
When a magnetic compass is brought close to a current carrying wire there is deflection in the compass due to the formation of magnetic field. The magnetic field lines around a long wire which carries an electric current form concentric circles around the wire. The direction of the magnetic field is perpendicular to the wire.
The deflection increases when the current is increased because the magnitude of the field is directly proportional to the strength of the electric current. That is, stronger the electric field, stronger the magnetic field.
A rectangular coil of single turn, having area A, rotates in a uniform magnetic field B with an angular velocity $$\omega$$ about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through $$90^o$$ is:
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$$\dfrac {\omega BA}{\pi}$$
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$$\dfrac {\omega BA}{2\pi}$$
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$$\dfrac {\omega BA}{4\pi}$$
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$$\dfrac {2\omega BA}{\pi}$$
Explanation
Initially flux, $$\phi=BA cos 0=BA$$
After rotating through an angle $$90^o$$.
Flux through the coil is zero.
So, $$\Delta \phi=BA$$
Angular speed $$=w$$, so, time period $$=\dfrac {2\pi}{\omega}=T$$
$$\dfrac {T}{4}$$ is time taken to rotate $$90^o$$.
So, $$\dfrac {\Delta \phi}{\Delta t}=\dfrac {BA}{T/4}=\dfrac {2BA\omega}{\pi}$$
If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the induced e.m.f. will be
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0.005 volt
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0.5 volt
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0.05 volt
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5 volt
Explanation
$$\varepsilon=L\times (rate \ of \ change \ of \ current)$$
$$=(5\times 10^{-3})(1/0.1)$$
$$=0.05V$$.
The SI unit of inductance, the henry can be written as
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weber/ ampere
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volt second/ ampere
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joule/ $$ampere^2$$
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all of the above
Explanation
$$ e = \dfrac{d\phi}{dt} = L\dfrac{dI}{dt} $$
$$\Rightarrow [volt] = \dfrac{[weber]}{[second]} = [L]\dfrac{[Ampere]}{[second]}$$
also,
Energy dissipated(E) $$ = VIt$$
$$\Rightarrow [Joule] = [volt][Ampere][second] \ Rightarrow [volt-second] = \dfrac{[Joule]}{[Ampere]}$$
so units of self inductance L is,
$$[L] = \dfrac{weber}{Ampere} = \dfrac{volt - second}{Ampere} = \dfrac{Joule}{Ampere^2}$$
A coil has 200 turns and area of $$70 cm^2$$. The magnetic field perpendicular to the plane of the coil is $$0.3 Wb/m^2$$ and take 0.1 sec to rotate through $$180^o$$. The value of the induced e.m.f. will be
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8.4 V
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84 V
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42 V
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4.2 V
Explanation
Change in flux $$=2BAN$$
$$\therefore$$ Induced e.m.f. $$=\dfrac {2\times 0.3\times 200\times 70\times 10^{-4}}{0.1}$$
$$=8.4V$$
According to Faraday's law of electromagnetic induction:
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electric field is produced by time varying magnetic flux
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magnetic field is produced by time varying electric flux
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magnetic field is associated with a moving charge
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none of these
Explanation
According to Faraday's Law of electromagnetic induction, if magnetic flux through a coil changes, it leads to an induced emf across the coil. Hence, electric field is produced by time-varying magnetic fields.
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