Question

electricity and magnetism

(2 pt) A loop of wire forming an equilateral triangle carries a steady current I. a. Find the magnetic field at the center of loop of wire forming a triangle which carries a steady current I. Let R be the closest distance from the center to the sides the triangle. b. Compare your result to the magnetic field at the center of circle of radius R. In which case is the field greater? Qualitatively explain why.

Answer 1

a)

The figure for the system is given below.

The total magnetic field at the centre is equal to the sum of the magnetic fields due to AB, BC and CA.

Due to the symmetry of the system (the triangle being equilateral), the magnetic fields due to AB, BC and CA are equal. So, to get the magnitude of the electric field, we can just find the field due to one of the sides and multiply it by 3.

Let's calculate the magnetic field due to side AB.

We know, the formula for the magnetic field due to a finite wire segment carrying current is given by the formula

Where

As the triangle is an equilateral triangle, we must have by symmetry

and also

Thus, from equation (1), we have, magnetic field due to one side of the triangle

Therefore, the total magnetic field at the centre

b)

We know that the magnetic field at the centre of a circle of radius R carrying current I is given by the formula

Dividing the two magnetic fields to compare

Therefore we see that the total magnetic field at the centre of the equilateral triangle is smaller than the magnetic field at the centre of a circle. So, the magnetic field at the centre of the circle is greater.

This result is expected because, for any given value of R, only the circle can have the longest length at the shortest distance from the centre and also the largest perimeter.

The magnitude of the magnetic field at a point is more when the distance between the wire and the point is smaller and also when the length of the perimeter is larger. Both of these are maximum for any value of R in case of a circle.