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Question: A wire carrying a current of \[4\,{\text{A}}\] is in the form of a circle. It is necessary to have a...

A wire carrying a current of 4A4\,{\text{A}} is in the form of a circle. It is necessary to have a magnetic field of induction 106T{10^{ - 6}}{\text{T}} at the centre. The radius is:
A. 2.51m2.51\,{\text{m}}
B. 25.1m25.1\,{\text{m}}
C. 251m251\,{\text{m}}
D. 0.251m0.251\,{\text{m}}

Explanation

Solution

First of all, we will find the expression of the magnetic field and substitute the required values in it. We will manipulate the expression accordingly and obtain the result.

Complete step by step answer:
In the given problem, we are supplied with the following data:
The amount of current that the wire is carrying is 4A4\,{\text{A}} .
The required magnetic field at the centre is 106T{10^{ - 6}}{\text{T}} .
The wire forms a circle.
We are required to form the radius of the wire loop.

For this we apply the formula which gives the magnetic field.
B=μ0i2rB = \dfrac{{{\mu _0}i}}{{2r}} …… (1)
Where,
BB indicates magnetic field.
μ0{\mu _0} indicates permeability constant.
ii indicates the amount of current flowing through the wire.
rr indicates the radius of the wire loop.

Now, substituting the required values in the equation (1), we get:

B=μ0i2r 106=4π×107×42r r=4π×107×4106 r=2.512m B = \dfrac{{{\mu _0}i}}{{2r}} \\\ \Rightarrow {10^{ - 6}} = \dfrac{{4\pi \times 1{0^{ - 7}} \times 4}}{{2r}} \\\ \Rightarrow r = \dfrac{{4\pi \times 1{0^{ - 7}} \times 4}}{{{{10}^{ - 6}}}} \\\ \Rightarrow r = 2.512\,{\text{m}} \\\

r2.51mr \sim 2.51\,{\text{m}}
Hence, the radius of the wire loop is 2.51m2.51\,{\text{m}} .
The correct option A.

Additional information:
A magnetic field is a vector field that determines the magnetic effect on electrical charges, electrical currents, and magnetised objects that pass. A force perpendicular to its own velocity and to the magnetic field is encountered by a charge travelling in a magnetic field.
In permanent magnets, which draw on magnetic materials such as iron and attract or repel other magnets, the effects of magnetic fields are frequently seen. Besides that, by affecting the motion of their outer atomic electrons, a magnetic field that varies with position can exert a force on a variety of non-magnetic materials.

Note: While solving this problem, we need to know about the magnetic field and its effects. The magnetic field depends on the current flowing through the wire, number of turns and material used.