Question

An infinite bent wire carrying a current = 3A lies in xy plane. One half of the wire is alinged along x axis, while the other half lies in xy plane forming an angle of 50 degree with x axis.

Answer 1

B_net=(μ₀3)/(4πa)√(2+2cos50°) directed at 115° from the +x–axis.

Answer 2

We will show one acceptable method to “sum” the contributions of the two semi‐infinite (straight) wires. (Any answer equivalent to the one below is correct.)

Step 1. (Field from a semi–infinite wire)

A well‐known result (which you may derive using the Biot–Savart law) is that the magnetic field at a point a perpendicular distance a from a semi–infinite straight wire is

Bₛ = (μ₀ I)/(4πa).

If a “full” infinite wire produces

B∞ = (μ₀ I)/(2πa),

then the field at the endpoint is exactly half that value.

Step 2. (The “bent” wire)

In the problem the infinite wire is bent at the vertex (which we take as the origin). One branch goes along the +x–axis; the other branch (still in the xy–plane) makes an angle of 50° with the x–axis. (Assume the currents in the two arms both go away from the vertex.)

Thus the magnetic field at the vertex is the vector sum of the fields due to the two semi–infinite wires. Their magnitudes are both

B₀ = (μ₀ I)/(4πa) with I = 3 A.

Let the direction of the field due to a straight current–element be determined by the right–hand rule (with the rule giving a field “circling” the wire). One may show that when one evaluates the contributions at the common end (vertex) the two fields are perpendicular to the respective wires. (That is, the field contribution from the branch along x–axis points in a direction which is perpendicular to the x–axis and similarly for the other branch.)

Thus, if we choose a coordinate system in which the “field directions” are defined by:

• For the x–axis branch, since the wire is along x, the field is directed perpendicular to x. (One may take this as along the “+y–direction” for definiteness.)
• For the second branch (making 50° with the x–axis) its field is perpendicular to it; that is, it is directed making an angle of 50°+90° = 140° with the x–axis.

It is most convenient to resolve these two fields into components.

Write

B₁ = (μ₀ I)/(4πa) (from the x–axis wire)

B₂ = (μ₀ I)/(4πa) (from the 50° branch)

The directions are:

• B₁ is along +y.
• B₂ makes an angle 140° with the +x–axis.

Thus, the components of B₂ are:

B₂ₓ = B₀ cos(140°) = –B₀ cos(40°),

B₂ᵧ = B₀ sin(140°) = B₀ sin(40°),

with B₀ = (μ₀ I)/(4πa).

Now the net field is

B_netₓ = B₁ₓ + B₂ₓ = 0 – B₀ cos40° = –B₀ cos40°,

B_netᵧ = B₁ᵧ + B₂ᵧ = B₀ + B₀ sin40° = B₀ (1 + sin40°).

Its magnitude is

|B_net| = B₀ √[ (cos40°)² + (1+ sin40°)² ].

Using numerical values (to 3–significant figures):

cos40° ≈ 0.766 and sin40° ≈ 0.643,

we have

|B_net| = B₀ √[ (0.766)² + (1+0.643)² ]

= B₀ √[ 0.587 + (1.643)² ]

= B₀ √[ 0.587 + 2.701 ]

= B₀ √[ 3.288 ]

≈ B₀ (1.815).

Since B₀ = (μ₀ I)/(4πa) with I = 3 A, we get

|B_net| ≈ (1.815 μ₀×3)/(4πa) = (5.445 μ₀)/(4πa).

The direction (angle measured from +x, say) is given by

tan θ = (net y–component)/(absolute net x–component)

= [ B₀ (1+ sin40°) ]/(B₀ cos40°)

= (1+ sin40°)/cos40°

≈ (1 + 0.643)/(0.766)

≈ 1.643/0.766 ≈ 2.145.

Thus, θ ≈ arctan(2.145) ≈ 65°.

But note: the net x–component is negative while the y–component is positive. Hence the vector lies in the second quadrant. So the angle measured from the +x axis is actually 180° – 65° = 115°.

A NOTE ON ALTERNATIVE APPROACHES

Many texts (and exam problems in JEE/NEET) give the result in “closed‐form” if the observation point is the vertex of a bent wire. In our problem one may also show (by a suitable change of variables) that the net field magnitude can be written in the form

B_net = (μ₀ I)/(4πa) √[2+2 cos50°].

A short calculation shows that

cos50° ≈ 0.643 → √[2+2×0.643] = √[2+1.286] ≈ √3.286 ≈1.813,

which is exactly the factor we obtained above.

One must then also take care to get the correct direction by vector‐addition; it turns out that one acceptable answer is to give the net field in magnitude and mention that its direction is at an angle of about 115° (measured from the +x–axis) in the xy–plane.

Final Answer

An acceptable answer is:

B_{net}=(μ_03)/(4πa)√(2+2cos50°) directed at 115° from the +x–axis.

(Any answer that is equivalent is correct.)

Summary of the Answer (minimal explanation)

1. The magnetic field at the end of a semi–infinite straight wire is B_{0}=(μ_0 I)/(4π a).
2. For the two branches (one along +x and one at 50° with +x), their fields (of equal magnitude B_0) are perpendicular to the wires.
3. Resolving the field from the 50° branch gives components: B_{2x}=-B_0cos40°, B_{2y}=B_0sin40°.
4. Adding the field from the x–axis branch (which is B_{1} = B_0 in the +y–direction) yields: B_{net}=√((B_0cos40°)^2+[B_0(1+sin40°)]^2).
5. This simplifies to B_{rm net}=(μ_03)/(4πa)√(2+2cos50°) with the net vector pointing in the second quadrant (115° from the +x–axis).