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A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its center at the origin. The magnetic induction at the center of the square loop is

Electrodynamics problems

Physics INK

Important to note that, The magnetic field of a n-side polygon carrying current I is given by 

`B=(nmu_0I)/(2pid)sin(pi/n)`

Where d is the perpendicular distance from the each side.

Problem: A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its center at the origin. The magnetic induction at the center of the square loop is: [GATE 2001]

(A)`(2sqrt2mu_0I)/(pia)hate_x`,    ✅(B)`(2sqrt2mu_0I)/(pia)hate_z`,     (C)`(2sqrt2mu_0I)/(pia^2)hate_z`,    (D)`(2sqrt2mu_0I)/(pia^2)hate_x`

Answer:

The square in the x-y pane so the magnetic field in z axis direction.

The magnetic field at the center of the square is 

`vecB=(4mu_0I)/(2pid)sin(pi/4)odot`

or, `vecB=(2sqrt2mu_0I)/(pia)odot`

or, `vecB=(2sqrt2mu_0I)/(pia)hate_z`

Option B  is correct.

Problem: A conducting wire is in the shape of a regular hexagon, which is inscribed inside an imaginary circle of radius R, as shown. A current I flows through the wire. The magnitude of the magnetic field at the centre of the circle is [JEST 2016]

(A)`(sqrt3mu_0I)/(2piR)`,    (B)`(mu_0I)/(2sqrt3piR)`,    ✅ (C) `(sqrt3mu_0I)/(piR)`,     (D) `(3mu_0I)/(2piR)`

Answer:

The perpendicular distance of the side and the center of the polygon is 

`d=Rsin(360^circ/(6xx2))`

or, `d=(sqrt3R)/2`

The magnetic field at the center of the hexagon is

`vecB=(6mu_0I)/(2pid)sin(pi/6)odot`

or, `vecB=(3mu_0I)/(pi)2/(sqrt3R)1/2odot`

or, `vecB=(sqrt3mu_0I)/(piR)`

Option C  is correct.

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