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Energy conservation - class-XII

Description: energy conservation
Number of Questions: 25
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Tags: electromagnetic induction and alternating currents physics electromagnetic induction
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A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of 40 rad $s^{-1}$ in a uniform horizontal magnetic field of magnitude $2 \times 10^{-2}$ T. If the coil form a closed loop of resistance 8 $\Omega$, then the average power loss

  1. $2.07\times 10^{-3}$W

  2. $1.23\times 10^{-3}$W

  3. $3.14\times 10^{-3}$W

  4. $1.80\times 10^{-3}$W


Correct Option: A
Explanation:
Here, $r = 6 cm = 6 \times 10^{-2}m,$ 
$N = 20, \omega = 40 \, rad \, s^{-1}$
$B = 2 \times 10^{-2} T, R = 8 \Omega$
Maximum emf induced, $\epsilon \, = \, NAB \omega$
$ = \, N(\pi r^2)B \omega$
$ = 20 \times \pi \times (6 \times 10^{-2})^2 \times 2 \times 10^{-2} \times 40 \, = \, 0.18 V$

Average value of emf induced over a full cycle $\epsilon _{av}\, = \, 0$ 

Maximum value of current in the coil, $ I \, = \, \dfrac{\varepsilon }{R} = \dfrac{0.18}{8} = 0.023 \,A$

Average power dissipated, $ P \, = \, \dfrac{\varepsilon I}{2} \, = \, \dfrac{0.18 \times 0.023}{2} \, = \, 2.07 \times 10^{-3} W$

The axle of a circular wheel of radius R is held horizontally by two identical strings of equal lengths separated by a distance D. The tension in each string is $T _0$. The rim of the wheel carries a total charge $+$Q distributed uniformly on it. The wheel is vertical and is kept in a uniform vertical magnetic field $\vec{B}$. It is now rotated at an angular speed $\omega$. If the string break at a tension of $3T _0/2$, than the maximum possible value of $\omega$ at which the wheel can be rotated without breaking a string is $\dfrac{DT _0}{QBR^2}$.

  1. True

  2. False


Correct Option: A

A conductor of $3\ m$ length is moving perpendicular to its length as well as a magnetic field of ${10}^{-3}\ T$ with a speed of ${10}\ m/s$, then the force required to move it with this constant speed is

  1. $0.3\ N$

  2. $0.9\ N$

  3. Zero

  4. $3 \times {10}^{-3}\ N$


Correct Option: C
Explanation:

Force on rod moving in a magnetic fileld is given by:

$F=BIL$
Where $I$ is the current in the rod.

According to the question;
$I=0$
Therefore,
$F=0$

Thus,
No force is exerted on the conductor in the magnetic field.
Thus, no external field is required to move the conductor at constant speed.

A horizontal straight conductor (otherwise placed in a closed circuit) along east-west direction falls under gravity. There is:-

  1. No induced electromotive force along the length

  2. No induced current along the length

  3. An induced current from west to east

  4. None of the above.


Correct Option: C

In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotated about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is:

  1. $\dfrac { B\pi { r }^{ 2 }\omega }{ 2R }$

  2. $\dfrac { { \left( B\pi { r }^{ 2 }\omega \right) }^{ 2 } }{ 2R }$

  3. $\dfrac { { \left( B\pi r\omega \right) }^{ 2 } }{ 2R }$

  4. $\dfrac { { \left( B\pi r{ \omega }^{ 2 } \right) }^{ 2 } }{ 8R }$


Correct Option: D

To produce a field of magnetic $\pi$ tesla at the center of circular loop of diameter 1 m, the current flowing through loop is :

  1. $5\times 10^6\;A$

  2. $10^7\;A$

  3. $2.5\times 10^6\;A$

  4. $2\times 10^6\;A$


Correct Option: C
Explanation:

Magnetic field due to circular loop at center O due to current $i$ is given as,

$B=\dfrac{\mu _0 I}{2r}=\dfrac{4\pi\times 10^{-7} \times I}{2\times 0.5}$
$\pi \times 1=4\pi \times 10^{-7} I$
$I=2.5\times 10^{6} A$

A metal ring of radius $r = 0.5\ m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2\ T$ carries a current $I = 100\ A$. The tension in newtons developed in the ring is

  1. $100$

  2. $50$

  3. $25$

  4. $10$


Correct Option: D

A magnet is brought towards a coil (i) speedly (ii) slowly, then the induced electromotive force charge will be respectively

  1. More in first case / More in first case

  2. More in first case / Equal in both case

  3. Less in first case / More in second case

  4. Less in first case / Equal in both case


Correct Option: B

An electric charge $+ q$ moves with velocity $\bar{V}=3\hat{i}+4\hat{j}+\hat{k}$, in an electromagnetic field given by $\bar{E}=3\hat{i}+\hat{j}+\hat2{k}$ $\bar{B}=\hat{i}+\hat{j}+\hat3{k}$.The y-component of the force experienced by +q is:

  1. $-7 q$

  2. $11 q$

  3. $5 q$

  4. $3 q$


Correct Option: A
Explanation:

Given that,

Velocity, $v=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)$

Electric field, $E=3\hat{i}+\hat{j}+2\hat{k}$

Magnetic field, $B=\hat{i}+\hat{j}+3\hat{k}$

Force,

$ F=q\left( \text{E+v}\times \text{B} \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+\left( (3\hat{i}+4\hat{j}+\hat{k})\times (\hat{i}+\hat{j}+3\hat{k} \right) \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+11\hat{i}-8\hat{j}-\hat{k} \right) $

$ F=q\left( 14\hat{i}-7\hat{j}+\hat{k} \right) $

So, y-component of force experienced by +q is

${{F} _{y}}=-7q$

A boat is moving due East in region where the earth's magnetic field is $5.0 \times {10^{ - 5}}\,N{A^{ - 1}}{m^{ - 1}}$ North and horizontal. The boat carries a vertical aerial $2 m$ long. if the speed of the boat is $1.50\,ms - 1,$ the magnitude of the induced emf in the wire of aerial is 

  1. $0.75 mV$

  2. $0.50 mV$

  3. $40.15 mV$

  4. $1 mv$


Correct Option: B

 A coil of circular cross-section having $100$ turns and $4 \mathrm { cm } ^ { 2 }$ face area is placed with its axis parallel to a magnetic field which decreases by $10 ^ { - 2 } \mathrm { Wb } \mathrm { m } ^ { - 2 }$ in $0.01 \mathrm { s }$. The e.m.f induced in the coil is: 

  1. $200 \mathrm { mV }$

  2. $0.4 \mathrm { mV }$

  3. $400 \mathrm { mV }$

  4. $4 \mathrm { mV }$


Correct Option: A

Radius of current carrying coil is 'R'. The ratio of magnetic field at a axial point which is R distance away from the centre of the coil to the magnetic field at the centre of the coil :-

  1. $\lgroup \frac{1}{2} \rgroup^{1/2}$

  2. $\frac{1}{2}$

  3. $\lgroup \frac{1}{2} \rgroup^{3/2}$

  4. $\frac{1}{4}$


Correct Option: C
Explanation:

$\begin{array}{l} \vec { B } \, at\, R\, dis\tan  ce=\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { { x^{ 2 } }+{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { 2{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2.{ { \left( 8 \right)  }^{ 1/2 } }.{ R^{ 3 } } } }  \ =\frac { { { \mu _{ o } }i } }{ { 4\sqrt { 2 } R } }  \ \vec { B } \, \, at\, \, centre=\frac { { { \mu _{ o } }i } }{ { 2R } }  \ Ratio=\frac { { { \mu _{ o } }i.2R } }{ { 4\sqrt { 2 } R.{ \mu _{ o } }i } } =\frac { 1 }{ { 2\sqrt { 2 }  } } ={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 3/2 } } \ Hence, \ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

The magnetic induction due to a magnet on the equatorial line at a distance 0.2 m is $54 \times 10^{-6}$T. The magnetic induction at 0.3m is

  1. $1.6 \times 10^{-6} \quad T$

  2. $1.6 \times 10^{-5} \quad T$

  3. $3.2 \times 10^{-6} \quad T$

  4. $3.2 \times 10^{-5} \quad T$


Correct Option: B

A uniform metal rod is moving with a uniform velocity $v$ parallel to a long straight wire carrying a current $I$. The rod is perpendicular to the wire with its ends at distances $r _{1}$ and $r _{2}$ with $(r _{2} > r _{1})$ from it. The emf induced in the rod is

  1. Zero

  2. $\dfrac {\mu _{0}Iv}{2\pi}\log _{e}\left (\dfrac {r _{2}}{r _{1}}\right )$

  3. $\dfrac {\mu _{0}Iv}{2\pi}\log _{e}\left (\dfrac {r _{1}}{r _{2}}\right )$

  4. $\dfrac {\mu _{0}Iv}{4\pi}\left (1 - \dfrac {r _{1}}{r _{2}}\right )$


Correct Option: B

If the permeability of iron piece is $3 \times 10 ^ { - 3 }$ and intensity of magnetising field of iron piece is 120 A/m, then what is the magnetic induction of iron piece 

  1. $0.36 \mathrm { Wb } / \mathrm { m } ^ { 2 }$

  2. $5 \times 10 ^ { - 3 } \mathrm { Wb } / \mathrm { m } ^ { 2 }$

  3. $40 \mathrm { Wb } / \mathrm { m } ^ { 2 }$

  4. $2.5 \times 10 ^ { - 4 } \mathrm { Wb } / \mathrm { m } ^ { 2 }$


Correct Option: A

A conducting ring of radius r is placed perpendicularly inside a time verying magnetic field given by $B={ B } _{ 0 }+\alpha t.{ B } _{ 0 }$ and $\alpha $ are positive constants. E. m. f induced in the ring is 

  1. $-\pi \alpha r$

  2. $-\pi \alpha { r }^{ 2 }$

  3. $-\pi { \alpha }^{ 2 }{ r }^{ 2 }$

  4. $-\pi { \alpha }^{ 2 }{ r }$


Correct Option: B

In the figure magnetic points into the plane of paper and the rod of length $l$ is moving in the field such that the bottom most point has a velocity $v _1$ and the topmost point has the velocity $V _2(V _2>V _1)$ The emf induced is given by 

  1. $Bv _1l$

  2. $Bv _2l$

  3. $\cfrac 1 2 B(v _2+v _1)l$

  4. $\cfrac 1 2 B(v _2-v _1)l$


Correct Option: C
Explanation:
Velocity of a point at a distance $x$ from the bottom is given by,

$v=v _1+(\dfrac{v _1-v _2}{l})x$

Potential difference on a small length at this distance is $e=Bx(dx)$

Therefore,

Total potential difference, $e=\int^l _0 Bvdx$

$e=\int^l _0 B(v _1+\dfrac{(v _2-v _1)x}{l})dx$

$=\dfrac{B(v _1+v _2)l}{2}$

A uniform magnetic field exists in region given by $\vec B = 3\hat i + 4\hat j + 5\hat k$. A rod of length $5m$ is placed along $y$ moved along $x-axis$ with constant speed $1m/sec$. Then induced e.m.f. in the rod will be:

  1. $zero$

  2. $25V$

  3. $20V$

  4. $15V$


Correct Option: B

A square coil of side 0.5$\mathrm { m }$ has movable side It is placed such that its plane is perpendicularuniform magnetic field of induction 0.2$\mathrm { T }$ . If all sides are allowed to move with a speed of 0.1for 4 sec outwards, average indaced emf is

  1. Zero

  2. 0.01$\mathrm { V }$

  3. 0.028$\mathrm { V }$

  4. 0.072$\mathrm { V }$


Correct Option: D

The amplitude of a magnetic field, which is part of a harmonic electromagnetic wave in vacuum, is  $\mathrm { B } _ { 0 } =510\mathrm { nT } .$  What is the amplitude of the electric field of the wave? 

  1. $140 \mathrm { NC } ^ { - 1 }$

  2. $153 \mathrm { NC } ^ { - 1 }$

  3. $163 \mathrm { NC } ^ { - 1 }$

  4. None of these


Correct Option: B

A  $1\mathrm { m }$  long conducting rod rotates with an angular frequency of  $400\mathrm { rad } \mathrm { s } ^ { - 1 }$  about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of  $0.5\mathrm { T }$  parallel to the axis exists everywhere.Calculate the e.m.f developed between the center and the ring?

  1. $100 \mathrm { V }$

  2. $101 \mathrm { V }$

  3. $102 \mathrm { V }$

  4. $103 \mathrm { V }$


Correct Option: A

A charged particle enters in a uniform magnetic field with velocity at an angle of ${ 60 }^{ o }$ with the magnetic field. The pitch of helical path is x, the radius of helix is 

  1. $\dfrac { x }{ 2\sqrt { 3 } x } $

  2. $\dfrac { 2x }{ \sqrt { 3 } \pi } $

  3. $\dfrac { \sqrt { 3 } x }{ \pi } $

  4. $\dfrac { \sqrt { 3 } x }{ 2\pi } $


Correct Option: D

When a charged particle is projected perpendicular to a magnetic field then the

  1. Velocity of the particle remain constant

  2. Momentum of the particle remain constant

  3. Kinetic energy of the particle remain constant

  4. Path of particle is straight line


Correct Option: C

A $0.1m$ long conductor carrying a current of $50A$ is perpendicular to a magnetic field of $1.21mT$. The mechanical power to move the conductor with a speed of $1m{s}^{-1}$ is

  1. $0.25mW$

  2. $6.25mW$

  3. $0.625W$

  4. $1W$


Correct Option: B
Explanation:

Mechanical power required $=$ electrical power expended due to the induced emf 

                                               $= (Blv) \times I \ =( 0.00121 \times 0.1 \times 1 ) \times 50\ =6.25 m W $

A non conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B=4t^2$ is switched on a time $t=0$. Mass of the ring is m and radius is R. The ring starts rotating after $2$ seconds. The coefficient of friction between the ring and the tablets.

  1. $\displaystyle\frac{4qmR}{g}$

  2. $\displaystyle\frac{2qmR}{g}$

  3. $\displaystyle\frac{8qR}{mg}$

  4. $\displaystyle\frac{qR}{2mg}$


Correct Option: C
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