Question

A closed circuit is in the form of regular hexagon of side 'a'. If the current carries 'I'. What is the magnetic induction (B) at the centre of the hexagonal.

⚠️Detailed solution is must⚠️. ​

Answer 1

Answer:

hoe....it...helps...

Explanation:

Answer 2

Answer:  μ. I.√3/ π.a

Given: Closed-circuit in form of Hexagon with side A carrying current (I)

To Find: Magnetic induction (B) at the center of Hexagon

Step-by-step explanation:

Suppose ABCDEF is the regular hexagon with side A carrying current (I) let O be the center of the hexagon.

Now, we know the formula for magnetic induction between two points.

B= μ. I. (sinα+sinβ)/ 4π.r

Since it’s a regular hexagon. Thus, α=β= 30°

Now, let’s find magnetic induction at O due to section AB

B(AB)= μ. I. (sin30°+sin30°)/ 4π.r

        =  μ. I. (sin30°)/ 2π.r         ……………… (1)

In the above equation, we need to find the value of r which is a perpendicular distance from center O to section AB.

Drop of perpendicular bisector OM to AB

Now, in triangle OAM

Tan30°= AM/OM

⇒  1/√3= a/2r

⇒  r= √3a/2                              …………….. (2)

From (1) and (2)

BAB= μ. I.sin30°/ 2π.(√3a/2)

BAB= μ. I/ 2π.√3a

Hence, the magnetic field is due to the whole hexagon at center O= 6. B(AB)

B(at the center of the hexagon)= 6.μ. I/ 2π.√3a

B(at the center of the hexagon)= μ. I.√3/ π.a

Project code:#SPJ2