Magnetics Ques 109
- A conducting circular loop is made of a thin wire has area $3.5 \times 10^{-3} m^{2}$ and resistance $10$ $ \Omega$. It is placed perpendicular to a time dependent magnetic field $B(t)=(0.4 T) \sin (0.5 \pi t)$. The field is uniform in space. Then the net charge flowing through the loop during $t=0$ $ s$ and $t=10$ $ ms$ is close to
(2019 Main, 9 Jan I)
(a) $6 $ $mC$
(b) $21$ $ mC$
(c) $7 $ $mC$
(d) $14 $ $mC$
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Answer:
Correct Answer: 109.(d)
Solution:
Formula:
Magnetic Moment Of A Current Carrying Loop:
- Since, the magnetic field is dependent on time, so the net charge flowing through the loop will be given as
$ Q=\frac{\text { change in magnetic flux, } \Delta \phi _B}{\text { resistance, } R} $
As, $\Delta \phi _B=\mathbf{B} \mathbf{A}=B A \cos \theta$
where, $A$ is the surface area of the loop and ’ $\theta$ ’ is an angle between $B$ and $A$.
Here, $\quad \theta=0 \Rightarrow \Delta \phi _B=B A$
$\therefore$ For the time interval, $t=0 ms$ to $t=10 ms$,
$ \begin{aligned} Q & =\frac{\Delta \phi _B}{R} \\ & =\frac{A}{R}\left(B _{f \text { at } 0.01 s}-B _{i \text { at } 0 s}\right) \end{aligned} $
Substituting the given values, we get
$ \begin{aligned} & =\frac{3.5 \times 10^{-3}}{10}[0.4 \sin (0.5 \pi)-0.4 \sin 0] \\ & =3.5 \times 10^{-4}(0.4 \sin \pi / 2) \\ & =1.4 \times 10^{-4} C=14 mC \end{aligned} $
BETA
