
If coefficient of self induction of a coil is 1H, an emf. of 1V is induced, if
a).Current flowing is 1 A
b).Current variation rate is 1 A/s
c).Current of 1 A flows for one second
d).None of these
Answer
585.6k+ views
Hint: Remember the relation between self inductance and flux. We will use this basic relation and apply certain changes in this basic relation to get a relation between the required parameters. Using this obtained relation, we will comment upon the final answer.
Complete answer:
Inductance refers to the tendency to oppose the current. The current flowing in a conductor faces an opposition, not in the form of resistance, when it changes from one value to another. The phenomena behind this opposition is called inductance.
We know, for a current I flowing through a conductor consisting of N turns and a magnetic flux of φ , the relation is written as:
$N\phi =LI$ ----(i)
Here, $\phi $ is not mentioned.
But when we differentiate eq. (i) w.r.t. time, we get:
$\dfrac{d}{dt}(N\phi )=\dfrac{d}{dt}(LI)$
$\Rightarrow N\dfrac{d}{dt}(\phi )=L\dfrac{d}{dt}(I)$ ------(ii)
We know that for an induced e.m.f. :
$V=N\dfrac{d\phi }{dt}$ ------(iii)
Putting equation (iii) in (ii),
$V=L\dfrac{dI}{dt}$ ------(iv)
Now,
According to question L = 1H & V = 1 Volt
Then clearly, $\dfrac{dI}{dt}=1$
The above equation can be expressed in words as the rate of change of current is equals to 1. The unit of rate of change of current is ampere per second, i.e. , A/s .
Hence, If the coefficient of self induction of a coil is 1H, an emf. of 1V is induced, if Current variation rate is 1 A.
Thus, the answer is option (b).
Additional Information:
Some of the important formulae to be remembered are:
$N\phi =LI$
$V=L\dfrac{dI}{dt}$
$D=\dfrac{E}{\varepsilon }$
$\mu =\dfrac{B}{H}$
$E=-\Delta V$
Note:
Do not confuse the relations $N\phi =LI$ and $Hl=NI$. Look carefully to find out what specific parameters are asked and apply the formula appropriately. Try to find a relation between the required parameters with the help of known relations, if there are no direct relations.
Complete answer:
Inductance refers to the tendency to oppose the current. The current flowing in a conductor faces an opposition, not in the form of resistance, when it changes from one value to another. The phenomena behind this opposition is called inductance.
We know, for a current I flowing through a conductor consisting of N turns and a magnetic flux of φ , the relation is written as:
$N\phi =LI$ ----(i)
Here, $\phi $ is not mentioned.
But when we differentiate eq. (i) w.r.t. time, we get:
$\dfrac{d}{dt}(N\phi )=\dfrac{d}{dt}(LI)$
$\Rightarrow N\dfrac{d}{dt}(\phi )=L\dfrac{d}{dt}(I)$ ------(ii)
We know that for an induced e.m.f. :
$V=N\dfrac{d\phi }{dt}$ ------(iii)
Putting equation (iii) in (ii),
$V=L\dfrac{dI}{dt}$ ------(iv)
Now,
According to question L = 1H & V = 1 Volt
Then clearly, $\dfrac{dI}{dt}=1$
The above equation can be expressed in words as the rate of change of current is equals to 1. The unit of rate of change of current is ampere per second, i.e. , A/s .
Hence, If the coefficient of self induction of a coil is 1H, an emf. of 1V is induced, if Current variation rate is 1 A.
Thus, the answer is option (b).
Additional Information:
Some of the important formulae to be remembered are:
$N\phi =LI$
$V=L\dfrac{dI}{dt}$
$D=\dfrac{E}{\varepsilon }$
$\mu =\dfrac{B}{H}$
$E=-\Delta V$
Note:
Do not confuse the relations $N\phi =LI$ and $Hl=NI$. Look carefully to find out what specific parameters are asked and apply the formula appropriately. Try to find a relation between the required parameters with the help of known relations, if there are no direct relations.
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