Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 4.1.1

Square loop

A square loop of side $a$ lies on the $x y$ -plane with its centre at the origin and carries a constant current $I$ . Using Biot–Savart's law on each of the four straight segments, find the magnetic field at the centre $O$ of the loop. Compare the result with the field at the centre of the circular loop of Problem 4.1.2 when $b = a /2$ .

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-5.

Try it yourself first.

Strategy

Biot–Savart's law on one straight side, from its midpoint distance $a /2$ to O. All four sides circulate the same way round O, so each gives the same vector along $a_{z}$ and the total is four times one side. The integral is that of Problem 4.1.1 with finite limits.

1. Symmetry

The four sides are carried into each other by rotations of $9 0^{\circ}$ about the $z$ -axis, which leave O fixed, so they contribute equal vectors at O. Each contribution is perpendicular to the plane of the loop (every $d l^{'} \times R$ with both vectors in the $x y$ -plane is along $\pm a_{z}$ ), and with the current circulating counter-clockwise seen from $+ z$ all four point along $+ a_{z}$ .

2. One side: the points and R

Take the side lying on $y^{'} = a /2$ , parallel to the $x$ -axis, carrying $I$ in the $- a_{x}$ direction (Figure 4.1.26). With $d l^{'} = - d x^{'} a_{x}$ , the vector from the element at $(x^{'}, a /2, 0)$ to O is $R = - x^{'} a_{x} - \frac{a}{2} a_{y}$ , and $R = x^{'2} + a^{2} /4$ . Then

d l^{'} \times R = - d x^{'} a_{x} \times (- x^{'} a_{x} - \frac{a}{2} a_{y}) = \frac{a}{2} d x^{'} a_{z} .

3. Integrate along the side

B_{1} = \frac{μ _{0} I}{4 π} \frac{a}{2} a_{z} \int_{- a /2}^{a /2} \frac{d x ^{'}}{( x ^{'2} + a ^{2} /4 ) ^{3/2}} = \frac{μ _{0} I}{4 π} \frac{a}{2} a_{z} [\frac{x ^{'}}{( a ^{2} /4 ) x ^{'2} + a ^{2} /4}]_{- a /2}^{a /2} = \frac{μ _{0} I}{4 π} \frac{a}{2} \cdot \frac{2 ( a /2 )}{( a ^{2} /4 ) ( a / 2 )} a_{z} = \frac{2 μ _{0} I}{2 π a} a_{z} .

In general a straight segment at the distance $d$ gives $μ_{0} I (sin θ_{2} - sin θ_{1}) / (4 π d)$ , with $θ_{1, 2}$ the angles under which its ends are seen from the foot of the perpendicular (Exercise 4.1.4). Here $d = a /2$ and $θ_{1, 2} = \mp 4 5^{\circ}$ , so $sin θ_{2} - sin θ_{1} = 2$ .

4. Four sides

Answer

B (O) = 4 B_{1} = \frac{2 2 μ _{0} I}{π a} a_{z} [T]

Home exercise 4.1.1. One side of the square seen from +z: the element at x' is at the distance R=\sqrt{x'^2+a^2/4} from O, and d\vec l'\times\vec R points out of the page for every element. The ends are seen from O under \theta_{1,2}=\mp45^\circ. — **Figure 4.1.26.** Home exercise 4.1.1. One side of the square seen from $+ z$ : the element at $x^{'}$ is at the distance $R = x^{'2} + a^{2} /4$ from O, and $d l^{'} \times R$ points out of the page for every element. The ends are seen from O under $θ_{1, 2} = \mp 4 5^{\circ}$ .

Check

Units: $μ_{0} I / a$ is T. Comparison: the circular loop of Problem 4.1.2 with $b = a /2$ gives $B = μ_{0} I /2 b = μ_{0} I / a$ . The ratio is $22 / π = 0.90$ : the inscribed circle, whose current is everywhere at the distance $a /2$ , produces about $10 %$ more field than the square, whose corners are further away ( $a / 2$ ). Numbers: for $a = 20$ cm and $I = 5$ A, $B = 22 \times 4 π \times 1 0^{- 7} \times 5/ (π \times 0.2) = 28.3$ μT, against $31.4$ μT for the circle of radius $10$ cm. Exercise 4.1.10 extends this to a point on the axis of the square.

Home exercise 4.1.1: Square loop