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Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From DC, R = V/I = 100/1 = 100 ohm. From AC, Z = V/I = 100/0.5 = 200 ohm. Using Z^2 = R^2 + (2*pi*f*L)^2, we find 200^2 = 100^2 + (2*pi*50*L)^2, which solves to L approximately 0.55 H.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $\dfrac { \mu _ { 0 } I } { 2 \pi ^ { 2 } R }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $0.004 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Induced emf = -d(phi)/dt. Change in flux = N * B * A * (cos(0) - cos(180)) = 1000 * 0.4*10^-4 T * 0.05 m^2 * 2 = 0.004 Wb. Emf = 0.004 / 0.1 = 0.04 V.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $10 ^ { - 7 } \times \frac { q \omega } { r }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $36.73\ mH$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mutual inductance M = (mu_0 * N1 * N2 * A) / l. With N1 = 5000, N2 = 700, A = pi * (0.04)^2, l = 0.5, M = (4*pi*10^-7 * 5000 * 700 * pi * 0.0016) / 0.5 = 0.04417 H = 44.17 mH.