A coil having $n$ turns and resistance $R\Omega $ is connected with a galvanometer of resistance $4R\Omega $ . This combination is moved in time $t$ seconds from a magnetic field ${w_1}\,webber$ to ${w_2}\,webber$ . The induced current in the circuit is:
(A) $ - \dfrac{{{w_1} - {w_2}}}{{5Rnt}}$
(B) $ - \dfrac{{n({w_2} - {w_1})}}{{5Rt}}$
(C) $ - \dfrac{{{w_1} - {w_2}}}{{Rnt}}$
(D) $ - \dfrac{{n({w_2} - {w_1})}}{{Rt}}$
Answer
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Hint: In order to answer this question, first we will apply the formula of induced current in terms of resistances that is connected in the circuit with a galvanometer, and then we will find the value of $E$ with respect to the given time and substitute it in the main formula of induced current to solve it.
Complete step by step solution:
As per the question, we have to find the induced current in the circuit, so we will apply the formula of current in terms of resistance, as resistance is given:
$I = \dfrac{E}{{R`}}$ ……eq(i)
where, $I$ is the current,
$R`$ is the resistance.
Now, the value of $E$ with respect to the given time:
$E = - \dfrac{{nd\phi }}{{dt}}$
where, $n$ is the number of turns of the coil.
Now, we will substitute the value of $E$ in eq(i):
$ \Rightarrow I = - \dfrac{n}{{R`}}.\dfrac{{d\phi }}{{dt}}$ …………eq(ii)
Given that ${w_1}\,and\,{w_2}$ are the values of the flux associated with one turn of the coil.
Thus, equation(ii) becomes:-
$I = - \dfrac{n}{{R`}}[\dfrac{{{w_2} - {w_1}}}{{{t_2} - {t_1}}}]$ ………..eq(iii)
Total resistance of the combination is:
$R` = R + 4R = 5R$
and substituting the values of $R`$ and $({t_2} - {t_1}) = t$ in eq(iii), we will get:
$
\therefore I = - \dfrac{n}{{5R}}(\dfrac{{{w_2} - {w_1}}}{t}) \\
\Rightarrow I = \dfrac{{ - n({w_2} - {w_1})}}{{5Rt}} \\
$
Therefore, the required induced current is $ - \dfrac{{n({w_2} - {w_1})}}{{5Rt}}$ .
Hence, the correct option is (B) $ - \dfrac{{n({w_2} - {w_1})}}{{5Rt}}$ .
Note:
When a bar magnet is pushed into a coil of insulated copper wire attached to a galvanometer, an induced current is created in the coil, which indicates a change in magnetic field. As a result, a deflection is produced by the galvanometer (say towards left).
Complete step by step solution:
As per the question, we have to find the induced current in the circuit, so we will apply the formula of current in terms of resistance, as resistance is given:
$I = \dfrac{E}{{R`}}$ ……eq(i)
where, $I$ is the current,
$R`$ is the resistance.
Now, the value of $E$ with respect to the given time:
$E = - \dfrac{{nd\phi }}{{dt}}$
where, $n$ is the number of turns of the coil.
Now, we will substitute the value of $E$ in eq(i):
$ \Rightarrow I = - \dfrac{n}{{R`}}.\dfrac{{d\phi }}{{dt}}$ …………eq(ii)
Given that ${w_1}\,and\,{w_2}$ are the values of the flux associated with one turn of the coil.
Thus, equation(ii) becomes:-
$I = - \dfrac{n}{{R`}}[\dfrac{{{w_2} - {w_1}}}{{{t_2} - {t_1}}}]$ ………..eq(iii)
Total resistance of the combination is:
$R` = R + 4R = 5R$
and substituting the values of $R`$ and $({t_2} - {t_1}) = t$ in eq(iii), we will get:
$
\therefore I = - \dfrac{n}{{5R}}(\dfrac{{{w_2} - {w_1}}}{t}) \\
\Rightarrow I = \dfrac{{ - n({w_2} - {w_1})}}{{5Rt}} \\
$
Therefore, the required induced current is $ - \dfrac{{n({w_2} - {w_1})}}{{5Rt}}$ .
Hence, the correct option is (B) $ - \dfrac{{n({w_2} - {w_1})}}{{5Rt}}$ .
Note:
When a bar magnet is pushed into a coil of insulated copper wire attached to a galvanometer, an induced current is created in the coil, which indicates a change in magnetic field. As a result, a deflection is produced by the galvanometer (say towards left).
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