Tag: physics

Questions Related to physics

When 100 volts d.c. is applied across solenoid a current of 1.0 amp flows in it. When 100 volts a.c. is applied across the same coil, the current drops to 0.5 amp. If the frequency of the a.c. source is 50 Hz the impedance and inductance of the solenoid are

  1. 200 ohm and 0.55 Henry

  2. 100 ohm and 0.86 Henry

  3. 200 ohm and 1.0 Henry

  4. 100 ohm and 0.93 Henry


Correct Option: A

The inductance of a solenoid 0.5 m long of cross-sectional area $420 cm^{2}$ and with $500$ turns is 

  1. $12.5\ mH$

  2. $1.25\ mH$

  3. $15.0\ mH$

  4. $0.12\ mH$


Correct Option: B

A current I flows in an infinity long wire with cross section in the from of a semicircular ring of radius R the magnitude of the magnetic induction along its axis is :- 

  1. $\dfrac { \mu _ { 0 } I } { 2 \pi R }$

  2. $\dfrac { \mu _ { 0 } \mathbf { I } } { 4 \pi R }$

  3. $\dfrac { \mu _ { 0 } I } { \pi ^ { 2 } R }$

  4. $\dfrac { \mu _ { 0 } I } { 2 \pi ^ { 2 } R }$


Correct Option: C
Explanation:

Let semicircular ring of radius $R$ as shown in figure 

An elementary length $dl$ cut for finding magnetic field,
So, 
$dl = Rd\theta $
So, elementary current along $dl$ is $di = \frac{i}{{\pi R}} \times Rd\theta  = \frac{{id\theta }}{\pi }$.
Now, you can see that elementary part of length is perpendicular upon $dB$.
so,
$dB = \frac{{{\mu _0}di}}{{2\pi R}}$

$ = \frac{{{\mu _0}di}}{{2{\pi ^2}R}}$


Now, magnetic filed along axis $B = \int\limits _0^\pi  {db.\sin \theta d\theta }  = 2dB$
Now, put $dB$ in here,
So, $B = \frac{{{\mu _o}i}}{{{\pi ^2}R}}$
Hence, 
Option $C$ is the correct answer.

What is inductance of a 25 cm long solenoid if it has 1000 turns an radius of its circular cross-section is 5 cm ?

  1. 0.04 H

  2. 0.02 H

  3. 0.8 H

  4. 0.1 H


Correct Option: B

A coil of mean area 500 $cm^2$ and having 1000 turns is held perpendicular to a uniform field of 0.4 gauss. The coil is turned through $180^o$ in $\frac{1}{10}$second. The average induced e.m.f. :-

  1. $0.04 V$

  2. $0.4 V$

  3. $4 V$

  4. $0.004 V$


Correct Option: A

The M.I. of a disc  about its diameter is $2$ units. Its M.I. about axis through a point on its rim in the plane of the disc is

  1. $4$ units

  2. $6$ units

  3. $8$ units

  4. $10$ units


Correct Option: A

A ring of radius r is uniformly charged with charge $q$ . If the ring is rotated about it's axis with angular frequency $\omega$, then the magnetic induction at its centre will be-

  1. $10 ^ { - 1 } \times \frac { \omega } { q r }$

  2. $10 ^ { - 7 } \times \frac { 9 } { \omega r }$

  3. $10 ^ { - 7 } \times \frac { r } { q \omega }$

  4. $10 ^ { - 7 } \times \frac { q \omega } { r }$


Correct Option: D
Explanation:

$\begin{array}{l} T=\dfrac { { 2\pi  } }{ w }  \ i=\dfrac { { qw } }{ { 2\pi  } }  \ B=\dfrac { { { \mu _{ 0 } }i } }{ { 2r } } =\dfrac { { { \mu _{ 0 } } } }{ { 2r } } \times \dfrac { { qw } }{ { 2\pi  } }  \ ={ 10^{ -7 } }\times \dfrac { { qw } }{ r }  \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$

The self inductance of a coil having $500$ turns is $50$ mH. The magnetic flux through the cross-sectional area of the coil while current through it is $8$mA is found to be?

  1. $4\times 10^{-4} Wb$

  2. $0.04$ Wb

  3. $4\mu$ Wb

  4. $40$m Wb


Correct Option: A
Explanation:
Given that,

Number of turns $N =500$

Self inductance $L=50\times10^{-3}\ H$

Current $I=8\times10^{-3}\ A$

The magnetic flux through an inductor is the self inductance of coil times the current through it.

The flux is 
$\phi=LI$

$\phi=50\times10^{-3}\times8\times10^{-3}$

$\phi=4\times10^{-4}\ Wb$

So, the magnetic flux is $4\times10^{-4}\ Wb$

Hence, A is correct option.

What is the mutual inductance of coil and solenoid if a solenoid of length $0.50\ m$ and with $5000$ turns of wire has a radius $4\ cm$ and a coil of $700$ turns is wound on the middle part of the solenoid?

  1. $44.17\ mH$

  2. $48.98\ mH$

  3. $34.34\ mH$

  4. $36.73\ mH$


Correct Option: A

The mutual inductance of an induction coil is 5 H. In the primary coil, the current reduces from 5 A to zero in $10^{-3} s$. What is the induced e.m.f. in the secondary coil?

  1. 2500 V

  2. 25000 V

  3. 2510 V

  4. zero


Correct Option: B
Explanation:

$EMF=L\dfrac { di }{ dt } $

$=5\times \dfrac { 5 }{ { 10 }^{ -3 } } $

$=25000V$