Irodov 3.227: Two Half Wires at Right Angles – Amit & Deepali Sinha Foundation

Problem Irodov 3.227

A very long wire carrying a current $I = 5.0$ A is bent through a right angle. Through the corner draw the line perpendicular to both straight parts of the wire. Find the magnetic induction at the point of that line a distance $l = 35$ cm from the corner.

The wire bent at a right angle at O, carrying I (sense chosen) in along one leg and out along the other, and the point P on the perpendicular through O at distance l. — The wire bent at a right angle at $O$ , carrying $I$ (sense chosen) in along one leg and out along the other, and the point $P$ on the perpendicular through $O$ at distance $l$ .

Given

$I = 5.0$ A
$l = 35$ cm
$\mu_0 = 4\pi\times10^{-7}$ H/m

Find

B

P

Solution

Step 1One leg: a half-infinite wire seen from the perpendicular at its end

Each leg runs from the corner $O$ to infinity, and $OP$ is perpendicular to it, so the foot of the perpendicular from $P$ to the leg is the leg’s end $O$ . Measure $x$ along the leg from $O$ . An element $dx$ is at distance $r = (x^2 + l^2)^{1/2}$ from $P$ , and the sine of the angle between $d\vect l$ and $\vect r$ is $l/r$ , so the Biot-Savart law gives

dB = \frac{\mu_0 I}{4\pi}\,\frac{l\,dx}{(x^2 + l^2)^{3/2}}

Every element of the leg gives $d\vect B$ the same direction, that of $d\vect l\times\vect r$ , normal to the plane containing the leg and $P$ . Integrating along the leg:

B_1 = \frac{\mu_0 I}{4\pi}\int_0^{\infty}\frac{l\,dx}{(x^2 + l^2)^{3/2}} = \frac{\mu_0 I}{4\pi}\left[\frac{x}{l\,(x^2 + l^2)^{1/2}}\right]_0^{\infty}

B_1 = B_2 = \frac{\mu_0 I}{4\pi l}

(1)

The second leg is the same situation turned through $90^\circ$ about $OP$ , so it gives the same magnitude.

Step 2The two fields are perpendicular

The field of each leg at P is normal to the plane through that leg and P, so B_1 lies along the other leg's direction and B_2 along the first's; their sum B is parallel to the bisector of the right angle (dashed, in the plane of the wire). — The field of each leg at $P$ is normal to the plane through that leg and $P$ , so $\vect B_1$ lies along the other leg’s direction and $\vect B_2$ along the first’s; their sum $\vect B$ is parallel to the bisector of the right angle (dashed, in the plane of the wire).

The field of the first leg is normal to the plane of that leg and $OP$ , which is the direction of the second leg, and vice versa. So $\vect B_1 \perp \vect B_2$ , and by Pythagoras with (1)

B = \left(B_1^2 + B_2^2\right)^{1/2} = \sqrt 2\,\frac{\mu_0 I}{4\pi l}

Numerically,

B = \sqrt 2\times\frac{4\pi\times10^{-7}\times5.0}{4\pi\times0.35}\ \text{T}

B = \frac{\mu_0}{4\pi}\,\frac{\sqrt 2\,I}{l} = 2.0\ \mu\text{T}

(2)

$\vect B$ is parallel to the plane of the wire, along the bisector of the right angle; for the current drawn (in along one leg, out along the other) it points along the bisector toward the inside of the right angle, the side the two legs span, and reversing the current reverses it.

Check

Straightening the wire, so that the legs are at $180^\circ$ to each other, puts the two half-wire fields parallel, and they add to $2\mu_0 I/(4\pi l) = \mu_0 I/(2\pi l)$ , the field of an infinite straight wire. The units: $\mu_0 I/l$ is in T $\cdot$ m/A times A/m, tesla.

Remark

Each half of a long wire supplies exactly half the full wire’s field at a point opposite its end, whatever happens to the other half; bending only turns the two halves’ fields relative to each other. For legs at an angle $\alpha$ to each other, the two equal fields are at $\pi - \alpha$ , and the total is $2\sin(\alpha/2)$ times one of them: $\sqrt 2$ here, $2$ for a straight wire ( $\alpha = \pi$ ), and $0$ for a hairpin folded back on itself ( $\alpha = 0$ ), whose currents cancel.