Loop current for a specified field at a jumper centre: 22-Elec-A7 Electromagnetics, May 2015, Q4

Question 4: Loop current for a specified field at a jumper centre (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. A flat “racetrack” loop: two 50 m straights 5 m apart, closed at each end by a semicircular jumper, hence of radius $R = 2.5\ \text{m}$; the target field is 10$^{-5}$ T at the centre of a jumper, to be computed within 5 %.

Given data
Quantity	Symbol	Value
Straight-section length	$L$	$50\ \text{m}$
Separation of the straights	$s$	$5\ \text{m}$
Jumper radius	$R = s/2$	$2.5\ \text{m}$
Required flux density at the jumper centre	$B$	$1.0\times10^{-5}\ \text{T}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}\ \text{H/m}$
Accuracy allowed	—	$5\ \%$

Find. The loop current $I$ that produces that flux density.

Figure 4.1 — plan view of the racetrack loop. P is the centre of the near jumper; every conductor contributes a field out of the page there.

Approach. Superpose the Biot–Savart contributions of the four conductors at the jumper centre. All four contributions are perpendicular to the plane of the loop and add with the same sign, so only magnitudes are needed; the 5 % tolerance is the licence to drop the distant jumper, which is verified rather than assumed.

Near jumper (a half circle centred on P). Every element of the arc is at the same distance $R$ and perpendicular to the radius, so the full-circle result is simply halved: $$B_{\text{arc}} = \frac{1}{2}\cdot\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R} = \frac{4\pi\times10^{-7}}{4(2.5)}\,I = 1.2566\times10^{-7}\,I\ \ \text{T}$$
Each straight section. P lies on the perpendicular dropped from the near end of each straight, at distance $R = 2.5\ \text{m}$. For a finite straight the standard result is $$B_{\text{str}} = \frac{\mu_0 I}{4\pi R}\left(\sin\theta_2 - \sin\theta_1\right)$$ with $\theta_1 = 0$ at the near end and $\sin\theta_2 = L/\sqrt{L^2+R^2} = 50/\sqrt{50^2+2.5^2} = 0.99875$ at the far end — essentially the semi-infinite value, because the straights are twenty times longer than they are far away.
Evaluate one straight. Substituting, $$B_{\text{str}} = \frac{4\pi\times10^{-7}}{4\pi(2.5)}(0.99875)\,I = 3.995\times10^{-8}\,I\ \ \text{T}$$ and there are two such straights, contributing $7.990\times10^{-8}\,I$ between them.
Justify dropping the far jumper. The other semicircle is a 7.85 m length of conductor some 50 m away, so its contribution is of order $\mu_0 I \ell/(4\pi r^2)$. Integrating Biot–Savart numerically over it gives $1.92\times10^{-10}\,I$, i.e. 0.09 % of the total — far inside the 5 % tolerance the question grants, which is precisely why that tolerance is stated.
Total field per ampere. All contributions are directed out of the plane of the loop at P (the current circulates one way, and P is inside the loop), so they add arithmetically: $$\frac{B}{I} = 1.2566\times10^{-7} + 2(3.995\times10^{-8}) = 2.0556\times10^{-7}\ \ \text{T/A}$$
Solve for the current. $$I = \frac{B}{B/I} = \frac{1.0\times10^{-5}}{2.0556\times10^{-7}} = \boxed{48.6\ \text{A}}$$ Including the far jumper would give 48.60 A instead of 48.65 A — a 0.1 % change, confirming that 48.6 A is correct to well within the required accuracy.

It is worth noticing how the contributions divide: the short semicircular jumper supplies 61 % of the field and the two 50 m straights only 39 % between them. Proximity beats length in magnetostatics, because the Biot–Savart integrand falls as $1/r^2$ while the useful length of a straight conductor accumulates only as its projection onto the angular variable. A crude “single infinite wire” estimate would have missed the answer by a factor of two and a half.

Final results
Quantity	Symbol	Value
Jumper radius	$R$	$2.5\ \text{m}$
Near-jumper contribution	$B_{\text{arc}}/I$	$1.257\times10^{-7}\ \text{T/A}$
Each straight section	$B_{\text{str}}/I$	$3.995\times10^{-8}\ \text{T/A}$
Far jumper (neglected)	$B_{\text{far}}/I$	$1.9\times10^{-10}\ \text{T/A}\ (0.09\ \%)$
Total field per ampere	$B/I$	$2.056\times10^{-7}\ \text{T/A}$
Required loop current	$I$	$48.6\ \text{A}$

Concept & Theory

Topic. Biot–Savart superposition for a planar “racetrack” current loop.

Key relations. Circular arc of angle $\phi$: $B = \mu_0 I\phi/(4\pi R)$, so a semicircle gives $\mu_0 I/(4R)$. Finite straight: $B = \mu_0 I(\sin\theta_2-\sin\theta_1)/(4\pi R)$, with the semi-infinite limit $\mu_0 I/(4\pi R)$.

Why this works. Biot–Savart is linear, so the field of any wire shape is the sum of the fields of its pieces. For a point inside a planar loop every piece contributes in the same normal direction, which turns a vector problem into a scalar sum. Choosing the field point at an arc centre is what makes the problem closed-form: the arc integrand becomes constant, and the two straights present their perpendicular feet exactly at that same point, so both standard formulas apply without any further geometry.

Common pitfall. Using the full-circle formula $\mu_0 I/2R$ for a semicircular jumper (a factor-of-two error), and using the symmetric finite-wire result $2\sin\theta_{\max}$ as though P were opposite the midpoint of each straight — it is opposite an end, so only one $\sin\theta$ term appears.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §7.2 (Biot–Savart law, worked straight-conductor and arc cases); Hayt & Buck, Engineering Electromagnetics, §7.1–7.2.

Question 4 of 8: Loop current for a specified field at a jumper centre

Question 4: Loop current for a specified field at a jumper centre (20 marks)