
A thin circular ring of the area $A$ is perpendicular to the uniform magnetic field of induction $B$. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $R$. When the ring is suddenly squeezed to zero areas, the charge flowing through the galvanometer is
(A) $\dfrac{{BR}}{A}$
(B) $\dfrac{{AB}}{R}$
(C) $ABR\;$
(D) $\dfrac{{{B^2}A}}{{{R^2}}}$
Answer
219k+ views
Hint: To solve this question we first need to understand some terms such as magnetic flux and emf. Magnetic flux is defined as the number of magnetic field lines that pass through a given closed surface. Emf is defined as electrical action which is produced by non-electrical sources.
Formula used:
The magnetic flux
$\phi = BA$
where $B$ is magnetic field induction and $A$is area.
The induced emf can be given a
$E = \dfrac{{\Delta \phi }}{{\Delta t}}$
Complete step-by-step solution:
The magnetic flux is defined as the number of the magnetic field lines that pass through a given closed surface, hence
$\phi = BA$
where $B$ is the magnetic field and $A$is area.
Hence the change in magnetic flux is given $\Delta \phi $.
$\Delta \phi = \Delta \left( {BA} \right)$
$ \Rightarrow \Delta \phi = B\Delta A = BA$ ………. $(1)$
Now the induced emf can be given a
$E = \dfrac{{\Delta \phi }}{{\Delta t}}$
Substituting the value from the equation $(1)$ we get that
$E = \dfrac{{BA}}{{\Delta t}}$
Now we know that the electric current is given as the ratio of potential difference to resistance according to ohm’s law but here instead of potential difference, we are substituting emf, hence
$i = \dfrac{E}{R}$
$ \Rightarrow i = \dfrac{{BA}}{{R\Delta t}}$ ………….$(2)$
Now we also know that the electric current is the rate of change of charges, hence equation $(2)$ can be rewritten as
$i = \dfrac{Q}{{\Delta t}} = \dfrac{{BA}}{{R\Delta t}}$
$\therefore Q = \dfrac{{BA}}{R}$
Hence the charge flowing through the galvanometer when the ring is squeezed so that the area is reduced zero is given as $Q = \dfrac{{BA}}{R}$.
Therefore option (B) is the correct answer.
Note: Here we had used the concept of magnetic flux which can also be defined as a measurement of the total magnetic field that passes through an area given. Hence magnetic flux is very useful as it describes the effects of the magnetic force on any object that occupies the area.
Formula used:
The magnetic flux
$\phi = BA$
where $B$ is magnetic field induction and $A$is area.
The induced emf can be given a
$E = \dfrac{{\Delta \phi }}{{\Delta t}}$
Complete step-by-step solution:
The magnetic flux is defined as the number of the magnetic field lines that pass through a given closed surface, hence
$\phi = BA$
where $B$ is the magnetic field and $A$is area.
Hence the change in magnetic flux is given $\Delta \phi $.
$\Delta \phi = \Delta \left( {BA} \right)$
$ \Rightarrow \Delta \phi = B\Delta A = BA$ ………. $(1)$
Now the induced emf can be given a
$E = \dfrac{{\Delta \phi }}{{\Delta t}}$
Substituting the value from the equation $(1)$ we get that
$E = \dfrac{{BA}}{{\Delta t}}$
Now we know that the electric current is given as the ratio of potential difference to resistance according to ohm’s law but here instead of potential difference, we are substituting emf, hence
$i = \dfrac{E}{R}$
$ \Rightarrow i = \dfrac{{BA}}{{R\Delta t}}$ ………….$(2)$
Now we also know that the electric current is the rate of change of charges, hence equation $(2)$ can be rewritten as
$i = \dfrac{Q}{{\Delta t}} = \dfrac{{BA}}{{R\Delta t}}$
$\therefore Q = \dfrac{{BA}}{R}$
Hence the charge flowing through the galvanometer when the ring is squeezed so that the area is reduced zero is given as $Q = \dfrac{{BA}}{R}$.
Therefore option (B) is the correct answer.
Note: Here we had used the concept of magnetic flux which can also be defined as a measurement of the total magnetic field that passes through an area given. Hence magnetic flux is very useful as it describes the effects of the magnetic force on any object that occupies the area.
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