
An air-cored solenoid with length \[~30\] cm, area of cross-section \[25\text{ }c{{m}^{2}}\] and number of turns \[500,\] carries a current of \[2.5\text{ }A.\] The current is suddenly switched off in a brief time of \[{{10}^{-3}}\text{ }s.\] How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Answer
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Hint: We know that electromagnetic induction is the production of electromotive force otherwise known as voltage across an electrical conductor where the magnetic field changes. The magnetic field surrounds every electric current. Fluctuating magnetic fields created around alternating currents.
Complete answer:
As we know , Faraday's Law is the equation that mathematically describes electromagnetic induction. It states that voltage (EMF) will be induced when there is a change in the magnetic environment of a coiled wire. Many ways were discovered by Faraday for this to happen. For example: by changing the magnetic field strength by moving a magnet over a coil of wire or by moving a coil of wire through a magnetic field. Faraday’s law of electromagnetic induction states that a current is induced in a conductor which is in a changing magnetic field. In accordance with Faraday's first law, any minute variation in the magnetic field of the coil will result in an emf which is getting induced in the coil. Here inside the solenoid and away from the ends is given by;
$B=\dfrac{{{\mu }_{0}}NI}{l}$ and $\Phi =\dfrac{{{\mu }_{0}}NIA}{l}.$ Thus, the total flux linkage is $N\Phi =\dfrac{{{\mu }_{0}}{{N}^{2}}IA}{l}.$
Since we know that the $|\varepsilon |=\dfrac{d}{dt}(N\Phi )$ which implies that the $|\varepsilon {{|}_{av}}=\dfrac{total~~change~~in~~flux}{total~~time}.$
On substitution we get; \[|\varepsilon {{|}_{av}}=\dfrac{4\pi \times {{10}^{-7}}\times 25\times {{10}^{-4}}}{0.3\times {{10}^{-3}}}\times {{\left( 500 \right)}^{2}}\times 2.5=6.5V\]
Note:
Note that the faraday's law considers how the changing magnetic fields can cause current to flow in wires. Lenz’s law in electromagnetic induction states that the direction of the induced current will be such that the magnetic field generated by the induced current will oppose the initial changing magnetic field which created it. The direction of current flow is determined using Fleming’s right-hand rule.
Complete answer:
As we know , Faraday's Law is the equation that mathematically describes electromagnetic induction. It states that voltage (EMF) will be induced when there is a change in the magnetic environment of a coiled wire. Many ways were discovered by Faraday for this to happen. For example: by changing the magnetic field strength by moving a magnet over a coil of wire or by moving a coil of wire through a magnetic field. Faraday’s law of electromagnetic induction states that a current is induced in a conductor which is in a changing magnetic field. In accordance with Faraday's first law, any minute variation in the magnetic field of the coil will result in an emf which is getting induced in the coil. Here inside the solenoid and away from the ends is given by;
$B=\dfrac{{{\mu }_{0}}NI}{l}$ and $\Phi =\dfrac{{{\mu }_{0}}NIA}{l}.$ Thus, the total flux linkage is $N\Phi =\dfrac{{{\mu }_{0}}{{N}^{2}}IA}{l}.$
Since we know that the $|\varepsilon |=\dfrac{d}{dt}(N\Phi )$ which implies that the $|\varepsilon {{|}_{av}}=\dfrac{total~~change~~in~~flux}{total~~time}.$
On substitution we get; \[|\varepsilon {{|}_{av}}=\dfrac{4\pi \times {{10}^{-7}}\times 25\times {{10}^{-4}}}{0.3\times {{10}^{-3}}}\times {{\left( 500 \right)}^{2}}\times 2.5=6.5V\]
Note:
Note that the faraday's law considers how the changing magnetic fields can cause current to flow in wires. Lenz’s law in electromagnetic induction states that the direction of the induced current will be such that the magnetic field generated by the induced current will oppose the initial changing magnetic field which created it. The direction of current flow is determined using Fleming’s right-hand rule.
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