A V-shaped wire

The first step is finding what $C(r)$ is in terms of $r$ because the problem is unsolvable otherwise. To do this, we imagine the shape as $α$ approaches $90^{∘}$ .

The bent shape of the wire becomes a straight wire, and $\tan(\frac{α}{2})$ becomes 1, so $B(r)=C(r)$ .

The strength of the magnetic field at a distance $r$ from a straight infinitely long wire can be measured using a form of the Biot-Savart law: $B=\frac{μ_{0}I}{2\pi r}$

Thus, $C(r)=\frac{μ_{0}I}{2\pi r}$ . We then substitute this expression back into $C(r)$ to obtain $B(r)=\frac{μ_{0}I}{2\pi r}\tan\frac{α}{2}$ .

Plugging in the given values, and we get that $B=8.284 \times 10^{-6}T$ .

Given: A V-Shaped Wire, with vertex at origin and the lower half and upper half making angle-α with the positive x-axis.

For the given wire configuration, the magnetic field at point ( -r , 0) is
    B(r) = C(r) * tan (α/2)

To find: the magnitude of the magnetic field in Teslas if α = 45°, the current is I = 1A (flowing from below the x-axis to above), and r = 1cm.

Solution: Let c be the position vector of the point ( -r ,0 ) say P.

Consider first the upper half of the wire.

Let l be the length from the origin to any point Q on the upper half of the wire.

Note: i , j , k are the unit vectors along the rectangular axes.

So l = l * cos α i + l * sin α j

Then dl = dl * cos α i + dl * sin α j

Now, let a be the vector pointing from Q to point P.

So, l + a = c ; implies a = c - l = -(r + l * cos α) i - l * sin α j

dl × a = r * sin α * dl k

According to Biot - Savart Law; we have

B(r) = (µ0/4π ) * I * ∫(dl × a)/|a|^3 with limits from l=0 to l = infinity ;

(Note: Denominator is mod(a) to the power of 3) (Note: I is the current flowing through the wire)

After solving the definite integral we get ;

B(r) = (µ0/4π ) * I * (1/r) * tan(α/2) k

Similarly for the wire below the x-axis, we get the same result.

Therefore net magnetic field at point P will be the resultant magnetic field due to both the halves.

So, net magnetic field is

B(r) = 2 * (µ0/4π ) * I * (1/r) * tan(α/2) k ;

Also, we get C(r) = (µ0/2π ) * I * (1/r) k

Now , putting r= 0.01m ; α = 45° ; I = 1A and µ0/4π = 10^(-7) H/m , we get |B(r)|= 8.284 * 10^(-6) Tesla

Alter: We can use the following formula for any straight wire of length l,

whose perpendicular distance from the point of consideration is d and

Ɵ1 and Ɵ2 are the angles made by the nearer and the farther ends

respectively with the perpendicular.

|B(r)| = (µ0/4π ) * (I/d) * (sin Ɵ2-sin Ɵ1).

For the direction of magnetic field, we can use the right hand thumb rule. In this case, for infinitely long wire ,

Ɵ2=90° and Ɵ1 = π/2 - α.

Direction of magnetic field is in positive - z direction for both the wires.

On solving for magnitude of lower and upper wire we will get the same magnitude. Therefore the resultant magnetic field will the addition of magnetic field due to two wires

So, on putting the values of I, d as r sin α, Ɵ1 and Ɵ2. We get

B(r) = 2 * (µ0/4π ) * I * (1/r) * tan(α/2) k

Now , putting r = 0.01m ; α = 45° ; I = 1A and µ0/4π = 10^(-7) H/m , we get |B(r)|= 8.284 * 10^(-6) Tesla .

The formula for a straight-line wire is given as $\frac {\mu_0*I}{2\pir}$ . This is equivalent to $\frac {\mu_0}{4\pi} \times \frac {2I}{r}$ .

This function is $C(r)$ . Substitution of the given parameters yields (10E-7)(2)(1 A)/(0.01 m).

Because the $\tan\frac{\alpha}{2}$ already takes into account the V-shape of the wire, multiplying $C(r)$ by $\tan 22.5$ yields the final answer of 8.28E-6.

Consider the case where $\alpha = 90$ . In this case, the wire will be straight. The magnitude of the magnetic field from a current carrying wire is $\frac{\mu_0I}{2\pi r} = 2\left(\frac{\mu_0}{4\pi}\right)\frac{I}{r}$ .

Therefore, in this case $B(0.01cm) = 2\times 10^{-7}\times 100 = 2\times 10^{-5} = C(0.01cm)\tan\left(\frac{90}{2}\right) \implies C(0.01cm) = 2\times 10^{-5}$ .

We can now put this value into the function when $\alpha = 45$ .

$B(0.01cm) = C(0.01cm) \times \tan(22.5) = 2\times 10^{-5} \times 0.41421356237 = \boxed{0.00000828 T}$

Note that in place of $45^\circ$ , if we had $\alpha= 90^\circ$ , the wire would become a straight line and we would have its magnetic field at point $(-r,0)$ equal to $B(r)= \frac{\mu_0 I}{2 \pi r}$ However, we have $\alpha= 45^\circ$ , and it is also known that $B(r)= C(r) \tan \left ( \frac{\alpha }{2} \right )$ From here we can show that $C(r)= \frac{\mu_0 I}{2 \pi r }$ Thus, we have $B(r)= C(r) \tan \left ( \frac{\pi }{8} \right )= \frac{\mu_0I}{2 \pi r} \tan \left ( \frac{\pi }{8} \right )$ Plugging the values, we obtain $B(r)= \boxed{8.28 \times 10^{-6} \text{Tesla} }$