Question 2 of 8: Magnetic Field of a Circular Current Loop

Question 2: Magnetic Field of a Circular Current Loop (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Horizontal circular loop, radius $R=5$ cm $=0.05$ m, current $I=2$ A, clockwise as viewed from above.

Find. (i) $\vec B$ at the centre (direction + magnitude). (ii) The axial distance $z$ where $B(z)=\tfrac12 B(0)$.

Approach. Use the right-hand rule for the loop's sense of circulation to fix the direction of $\vec B$, the standard centre-of-loop formula for its magnitude, and the on-axis Biot–Savart result $B(z)=\mu_0IR^2/2(R^2+z^2)^{3/2}$ to solve for $z$.

1. Direction at the centre. Curling the right-hand fingers in the current's sense as seen by an observer above the loop (clockwise, looking downward, i.e. looking in $-\hat z$) points the thumb downward. So $\vec B(0)$ points vertically downward (away from the observer above, through the loop).
2. Magnitude at the centre. $$B(0)=\frac{\mu_0 I}{2R}=\frac{(4\pi\times10^{-7})(2)}{2(0.05)}$$ $$B(0)=\boxed{2.51\times10^{-5}\ \text{T} = 25.1\ \mu\text{T}}$$
3. On-axis field. At a distance $z$ from the centre along the axis, $$B(z)=\frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}.$$ Setting $B(z)=\tfrac12 B(0)=\dfrac{\mu_0 I}{4R}$ and dividing by $B(0)$: $$\frac{R^3}{(R^2+z^2)^{3/2}}=\frac12 \;\Rightarrow\; (R^2+z^2)^{3/2}=2R^3 \;\Rightarrow\; R^2+z^2=2^{2/3}R^2.$$
4. Solve for $z$. $$z=R\sqrt{2^{2/3}-1}=(0.05)\sqrt{2^{2/3}-1}$$ $$z=\boxed{3.83\ \text{cm}}$$ By the symmetry of $B(z)$ (it depends only on $z^2$), this value of the field also occurs at $z=-3.83$ cm — i.e. two points, one on each side of the loop's plane.