Question

The electric current in a circular coil of two turns produced a magnetic induction of 0.2 T at its centre. The coil is unwound and is rewound into a circular coil of four turns. The magnetic induction at the centre of the coil now is, in T:
(if same current flows in the coil)
(A) 0.2
(B) 0.4
(C) 0.6
(D) 0.8

Answer 1

Hint: It is known that there are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object. The presence of a charged object near a neutral conductor will force (or induce) electrons within the conductor to move. An effective induction ensures that new staff can quickly learn the University's policies, processes and practices. The term 'induction' is generally used to describe the whole process whereby employees adjust or acclimatise to their jobs and working environment.

 Complete step by step answer
Let us begin with Faraday's Principle of Electromagnetic Induction states that the emf induced in a loop due by a changing magnetic flux is equal to the rate of change of the magnetic flux threading the loop. The magnetic flux threading a coil of wire can be changed by moving a bar magnet in and out of the coil. For example, an electric generator produces a current because of electromagnetic induction. Other uses for electromagnetic induction include electric motors used in anything from washing machines to trains, electric hobs and cookers, transformers, welding and guitar pickups.
When there are two turns in the coil, then
$\mathrm{l}=2 \times 2 \pi \mathrm{r}_{1}$ or $\mathrm{r}_{1}=\dfrac{1}{4 \pi}$ then $\mathrm{B}_{1}=\dfrac{\mu_{\mathrm{o}} \mathrm{N}_{1} \mathrm{I}}{2 \mathrm{r}_{1}}=\dfrac{\mu_{\mathrm{o}} \times 2 \times \mathrm{I}}{2 \times(\mathrm{l} / 4 \pi)}=\dfrac{\mu_{\mathrm{o}} 4 \pi \mathrm{I}}{\mathrm{l}}$
When there are four turns in the coil, then
$\mathrm{l}=4 \times 2 \pi \mathrm{r}_{2}$ or $\mathrm{r}_{2}=\dfrac{1}{8 \pi}$
Then $\mathrm{B}_{2}=\dfrac{\mu_{\mathrm{o}} \mathrm{N}_{2} \mathrm{I}}{2 \mathrm{r}_{2}}=\dfrac{\mu_{\mathrm{o}} \times 4 \times \mathrm{I}}{2 \times(\mathrm{l} / 8 \pi)}=\dfrac{\mu_{\mathrm{o}} 16 \pi \mathrm{I}}{\mathrm{l}}$
So now we can write that:
$\dfrac{\mathrm{B}_{1}}{\mathrm{B}_{2}}=\dfrac{4}{16}=\dfrac{1}{4}=$ or $\mathrm{B}_{2}=4 \mathrm{B}_{1}=4 \times 0.2 \mathrm{T}=0.8 \mathrm{T}$

So, the correct answer is option D.

 Note: We can conclude that an AC (alternating current) generator utilizes Faraday's law of induction, spinning a coil at a constant rate in a magnetic field to induce an oscillating emf. The coil area and the magnetic field are kept constant, so, by Faraday's law, the induced emf is given by: If the loop spins at a constant rate. When an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it. The induced magnetic field inside any loop of wire always acts to keep the magnetic flux in the loop constant. Faraday's law of induction is one of the important concepts of electricity. It looks at the way changing magnetic fields can cause current to flow in wires. Basically, it is a formula/concept that describes how potential difference (voltage difference) is created and how much is created.