
A long solenoid with 15 turns per cm has a loop of area $2.0\text{ c}{{\text{m}}^{2}}$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 sec, calculate the induced emf in the loop while the current is changing.
Answer
492.3k+ views
Hint: We know by Faraday's law that a changing current produces changing magnetic fields and this produces a change in magnetic flux. A change in magnetic flux will produce an induced emf. which will oppose the changing flux. A solenoid consists of a piece of a conducting wire wrapped on it into a number of coils. It is used to generate magnetic fields with the help of current.
Complete step-by-step solution:
In this question, we need to find the induced emf due to a solenoid and we know that
The emf induced is given as
$e=\dfrac{d\phi }{dt}$
And $\phi =BA$
And the magnetic field (B) = ${{\mu }_{0}}NI$
Therefore, by combining all the equations above, we get
Induced Emf (e) = $\dfrac{d}{dt}(BA)$
Or $e=A{{\mu }_{0}}N\dfrac{di}{dt}$ ………….. (1)
Now, we have been given in the question that there are 15 turns per cm
Or we can say that there will be 1500 turns in one meter
Now, we have been given that the current in the solenoid changes from 2.0 A to 4.0 A in just 0.1 second.
This means we have $di=\text{ 2 A}$and $dt\text{ = 0}\text{.1 seconds}$
Area of the loop is given as (A) = $2.0\text{ c}{{\text{m}}^{2}}$ or $2\times {{10}^{-4}}\text{ }{{\text{m}}^{2}}$
Now, putting all the values above in equation (1), we get
$e=A{{\mu }_{0}}N\dfrac{di}{dt}$
Where ${{\mu }_{0}}=4\pi \times {{10}^{-7}}\text{ H/m}$
$\Rightarrow e=(2\times {{10}^{-4}})(4\pi \times {{10}^{-7}})(1500)\dfrac{2}{0.1}$
$\Rightarrow e=7.54\times {{10}^{-6}}\text{ V}$
Hence, we can say that the induced emf $e=7.54\times {{10}^{-6}}\text{ V}$
Note: Solenoids are primarily to produce a uniform magnetic field. Solenoids are used in research or R&D labs and also used in college laboratories. If we use permanent magnets, we know that we observe curved field lines. Solenoids are also special because their magnetism can be turned on or off with the help of an external supply while this is never achievable if we were using permanent magnets.
Complete step-by-step solution:
In this question, we need to find the induced emf due to a solenoid and we know that
The emf induced is given as
$e=\dfrac{d\phi }{dt}$
And $\phi =BA$
And the magnetic field (B) = ${{\mu }_{0}}NI$
Therefore, by combining all the equations above, we get
Induced Emf (e) = $\dfrac{d}{dt}(BA)$
Or $e=A{{\mu }_{0}}N\dfrac{di}{dt}$ ………….. (1)
Now, we have been given in the question that there are 15 turns per cm
Or we can say that there will be 1500 turns in one meter
Now, we have been given that the current in the solenoid changes from 2.0 A to 4.0 A in just 0.1 second.
This means we have $di=\text{ 2 A}$and $dt\text{ = 0}\text{.1 seconds}$
Area of the loop is given as (A) = $2.0\text{ c}{{\text{m}}^{2}}$ or $2\times {{10}^{-4}}\text{ }{{\text{m}}^{2}}$
Now, putting all the values above in equation (1), we get
$e=A{{\mu }_{0}}N\dfrac{di}{dt}$
Where ${{\mu }_{0}}=4\pi \times {{10}^{-7}}\text{ H/m}$
$\Rightarrow e=(2\times {{10}^{-4}})(4\pi \times {{10}^{-7}})(1500)\dfrac{2}{0.1}$
$\Rightarrow e=7.54\times {{10}^{-6}}\text{ V}$
Hence, we can say that the induced emf $e=7.54\times {{10}^{-6}}\text{ V}$
Note: Solenoids are primarily to produce a uniform magnetic field. Solenoids are used in research or R&D labs and also used in college laboratories. If we use permanent magnets, we know that we observe curved field lines. Solenoids are also special because their magnetism can be turned on or off with the help of an external supply while this is never achievable if we were using permanent magnets.
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