Tag: magnetic flux

Questions Related to magnetic flux

The magnetic flux density at a point distant $d$ from a long straight current carrying conductor is $B$, then its value at distance $d/2$ will be:

  1. $4B$

  2. $2B$

  3. $B/2$

  4. $B/4$


Correct Option: B
Explanation:

Flux density, B $=\dfrac{\phi}{A}$
$B _A=\dfrac{\mu _0 I}{2\pi d}=B$
$=\dfrac{\phi}{A}$
$B _A=\dfrac{\mu _0 I}{2\pi \dfrac{d}{2}}=2\dfrac{\mu _0 I}{2\pi d}$
$B _B=2\times B$
$B _B=2B$

The magnetic needle of a tangent galvanometer is deflected at an angle $30$ due to a magnet. The horizontal component of earth's magnetic field $0.34\times 10^{-4}T$ is along the plane of the coil. The magnetic intensity is:

  1. $1.96\times 10^{-4}T$

  2. $1.96\times 10^{-5}T$

  3. $1.96\times 10^{4}T$

  4. $1.96\times 10^{5}T$


Correct Option: B
Explanation:

The correct option is B.


Given,

$B=0.34\times10^{-4}T$

Deflected angle$\theta=30^0$

So magnetic intensity is $Btan 30^0$

$=0.34\times10^{-4}T\times\dfrac{1}{\sqrt3}$

$=1.96\times10^{-5}T$

Where,$tan 30^0=\dfrac{1}{\sqrt3}$, and $\sqrt3=1.73$c

A sphere of radius $R$ and charge $Q$ is placed inside an imaginary sphere of radius $2R$. Whose center coincides with the given sphere. The flux related to the imaginary sphere is:

  1. $\dfrac {Q}{\in _{0}}$

  2. $\dfrac {Q}{2\in _{0}}$

  3. $\dfrac {4Q}{\in _{0}}$

  4. $\dfrac {2Q}{\in _{0}}$


Correct Option: A

The flux linked with a coil changes with time according to the equation $\phi$ = a$t^2$ +bt +c. Then SI unit of a is 

  1. Volt

  2. Volt/sec

  3. Volt.sec

  4. Weber


Correct Option: B
Explanation:

The unit of $\phi$ is Volt-sec.

Now $\phi=a t^{2}+bt+c$
To meet the dimension requirement $at^{2}$ must be Volt-sec.
$a$ must Volt/sec.

In a circuit a coil of resistance $2\,\Omega$, then magnetic flux charges from $2.0\,Wb$ to $10.0\,Wb$ in $0.2\ sec.$ The charge flow in the coil during this time is:

  1. $5.0\ C$

  2. $4.0\ C$

  3. $1.0\ C$

  4. $0.8\ C$


Correct Option: B
Explanation:

The relation between the rate of change of charge (or current) and the flux is given by the following relation:  

$ \because \dfrac{dQ}{dt}=-\dfrac{1}{R}\dfrac{d\phi }{dt} $

$ \dfrac{dQ}{dt}=\dfrac{-(10-2)}{2}=4\,C $


Two coils $A$ and $B$ are wound on the same iron  core as shown in figure. The number of turns in the coil $A$ and $B$ are $N _{A}$ and $N _{B}$ respectively. Identity the correct statement 

  1. Both the coils have same magnitude of magnetic flux

  2. The magnetic flux linked are in the ratio $\dfrac{\phi A}{\phi B}=\dfrac{N _{A}}{N _{B}}$

  3. The induced emf across each coil are in the ratio $\dfrac{E _{A}}{E _{B}}=\left(\dfrac{N _{4}}{N _{B}}\right)^{2}$

  4. Both the coils have same magnitude of induced emf


Correct Option: B

The magnetic flux through a stationary loop with resistance R varies during the interval of time T as $\phi  = at(T - t)$ ./ The heat generated during this time neglecting the inductance of the loop will be :

  1. $\dfrac{{{a^2}{T^3}}}{{3R}}$

  2. $\dfrac{{{a^2}{T^2}}}{{3R}}$

  3. $\dfrac{{{a^2}T}}{{3R}}$

  4. $\dfrac{{{a^3}{T^3}}}{{3R}}$


Correct Option: C

A circular disc of radius $0.2$m isplaced in a uniform magnetic field of induction $\dfrac{1}{\pi}\left(\dfrac {Wb}{m^2}\right)$ in such a way that its axis ,makes an angle of $60^o$ with $\xrightarrow {B}$. The magnetic flux linked with the disc is

  1. $0.01$ Wb

  2. $0.02$ Wb

  3. $0.06$ Wb

  4. $0.08$ Wb


Correct Option: A

The magnetic flux through a coil is $4\times 10^{-4} W/b/m^2$ at time $t=0$.It reduces to $10\%$ of its original value in 't' seconds.If the induced e.m.f is $0.72 m V,$ then the time t is:

  1. $0.25 s$

  2. $0.05 s$

  3. $0.75 s$

  4. $1 s$


Correct Option: B
Explanation:

As given in question

magnetic turn $=(\phi _1) = 4 \times 10^{-4}wb/m^2$
at $t = 0$
At, $t =t _2$, the turn reduces to $10\%$, means 
$\phi _2 = 0.9\ \phi _1$
As, per the farady's law,
In duced Emf $= \dfrac{Nd\ \phi}{dt}$
$e = \dfrac{nd\phi}{dt}$      ...(1)
$e = 0.72mu$ pur in (1), take $N = 1$ turns are constant
$0.72 \times 10^{-3} = \dfrac{-(\phi _2-\phi _1)}{(t _2-t _1)}$
$0.72\times 10^{-3} = \dfrac{-(0.9-2)\phi _1}{(t _2-0)}$
$t _2 = \dfrac{(0.1)\times (4\times 10^{-4})}{(0.72\times 10^{-3})}$
$t _2 = 0.05\ sec$

In a uniform electric field $\vec {E}$ an imaginary cube of edge length $a$ is considered as shown. The outward flux linked with cube surface will be :

  1. $Ea^{2}$

  2. $\sqrt{2}Ea^{2}$

  3. $\sqrt{3}Ea^{2}$

  4. $2Ea^{2}$


Correct Option: A