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Question: In the given figure, the magnetic induction at the point \(O\) is ![](https://www.vedantu.com/ques...

In the given figure, the magnetic induction at the point OO is

A. μ0I4πr\dfrac{{{\mu _0}I}}{{4\pi r}}
B. μ0I2πr+μ0I4r\dfrac{{{\mu _0}I}}{{2\pi r}} + \dfrac{{{\mu _0}I}}{{4r}}
C. μ0I4πr+μ0I4r\dfrac{{{\mu _0}I}}{{4\pi r}} + \dfrac{{{\mu _0}I}}{{4r}}
D. μ0I4πr+μ0I4r - \dfrac{{{\mu _0}I}}{{4\pi r}} + \dfrac{{{\mu _0}I}}{{4r}}

Explanation

Solution

In magnetism, magnetic field is the space around an accelerated charge or current carrying conductor where magnetic force can be experienced, in order to find net magnetic induction at point O we will use the general formula of finding magnetic field due to current carrying wire.

Formula used:
Magnetic field due to semicircular arc of wire at centre having current II and radius rr is calculated as,
Bcentre=μ0I4r{B_{centre}} = \dfrac{{{\mu _0}I}}{{4r}}
Magnetic field due to wire at one end perpendicular to the wire at a distance of rr is calculated by,
Bend=μ0I4πr{B_{end}} = \dfrac{{{\mu _0}I}}{{4\pi r}}
Magnetic field due to wire at a point on the line of wire at any distance through it on any end is always zero.

Complete step by step answer:
According to the given diagram, we can see that the part of wire AB having current I and point O lies on the line of wire AB hence the magnetic field due to this part will be zero.
BAB=0{B_{AB}} = 0
Now, part of wire BCD is a semi-circular arc having radius of r and current I and centre at O then by using the formula Bcentre=μ0I4r{B_{centre}} = \dfrac{{{\mu _0}I}}{{4r}} the magnetic field at point O will be
BBCD=μ0I4r{B_{BCD}} = \dfrac{{{\mu _0}I}}{{4r}}

Now, the part of wire DE having current I, point O lies perpendicular to the wire at one end and having at a distance of r will have magnetic field according to the formula Bend=μ0I4πr{B_{end}} = \dfrac{{{\mu _0}I}}{{4\pi r}} so,
BDE=μ0I4πr{B_{DE}} = \dfrac{{{\mu _0}I}}{{4\pi r}}
Since magnetic field due to each individual part is same hence, total magnetic field at point O can be written as
Bnet=BBCD+BDE+BAB{B_{net}} = {B_{BCD}} + {B_{DE}} + {B_{AB}}
On putting the values we get,
Bnet=μ0I4πr+μ0I4r\therefore {B_{net}} = \dfrac{{{\mu _0}I}}{{4\pi r}} + \dfrac{{{\mu _0}I}}{{4r}}

Hence, the correct option is C.

Note: It should be remembered that, if the current in the wire is moving in anticlockwise direction then direction of magnetic field will be out of the plane and if current were flowing in clockwise direction then magnetic field direction would be into the plane.μ0{\mu _0} Is known as the permeability of free space.