On-Axis Field of a Clockwise Current Loop: 04-BS-9, May 2017, Q3

Question 3: On-Axis Field of a Clockwise 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.

Quantity	Value
Current $I$	$2$A
Loop radius $a$	$5$cm $=0.05$m
Field point height $z$	$5$cm above centre (so $z=a$)
Current sense (viewed from above)	clockwise

Find. The magnitude and direction of $\vec B$ at the field point.

Loop of radius $a$ carries current $I$ circulating clockwise viewed from above; the field point sits on the axis at height $z=a$.

Approach. Apply the standard on-axis field of a circular loop with $z=a$; determine the SIGN by the right-hand rule applied to the stated clockwise-from-above current sense.

Magnitude, on-axis loop field with $z=a$. $$B=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}=\frac{\mu_0Ia^2}{2(2a^2)^{3/2}}=\frac{\mu_0I}{4\sqrt2\,a}$$ $$B=\frac{(4\pi\times10^{-7})(2)}{4\sqrt2\,(0.05)}=\boxed{8.886\times10^{-6}\ \text{T}}$$
Direction by the right-hand rule. Curling the right-hand fingers in the current's sense as seen from above (clockwise) points the thumb DOWNWARD, i.e. the loop's magnetic moment $\vec m$ points down through the loop; the on-axis field is directed along $\vec m$ at every point on the axis: $$\boxed{\vec B=8.886\times10^{-6}\ \text{T, directed DOWNWARD (toward the loop), along the axis}}$$

Quantity	Result
Magnitude $B$	$8.886\times10^{-6}$ T
Direction	downward along the axis (toward the loop), set by the clockwise-from-above current

Concept & Theory

Topic. On-axis magnetic field of a circular current loop (Biot–Savart), and the right-hand-rule sign convention linking current sense to field direction.

Key relations. $B(z)=\mu_0Ia^2/[2(a^2+z^2)^{3/2}]$; at $z=a$ this simplifies to $\mu_0I/(4\sqrt2\,a)$.

Why this works. The on-axis field of ANY planar current loop points along the loop's magnetic moment $\vec m=I\vec A$ for every point on the axis, not merely in the far-field limit; a clockwise sense viewed from above corresponds to $\vec m$ pointing down (the right-hand-rule curl), which is why the field at a point ABOVE the loop still points DOWN, toward the loop.

Common pitfall. Defaulting to the more familiar counter-clockwise/upward-moment convention without checking the problem's stated viewing sense, which would flip the reported direction (though not the magnitude).

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart law, on-axis field of a circular loop, right-hand rule).

Question 3 of 8: On-Axis Field of a Clockwise Current Loop

Question 3: On-Axis Field of a Clockwise Current Loop (20 marks)