Question

28.33 calculate the magnitude of the magnetic field at point p of fig. e28.33 in terms of r, i₁, and i₂. what does your expression give when i₁ = i₂?

Explanation:

Step1: Magnetic field due to straight wires

The magnetic field at point \(P\) due to each of the two straight - wire segments (using the formula for the magnetic field due to an infinite straight wire \(B=\frac{\mu_0I}{2\pi r}\), but for semi - infinite wires, the contribution of a semi - infinite wire at a perpendicular distance \(r\) from the end is \(B = \frac{\mu_0I}{4\pi r}\)). Here, the two straight - wire segments (upper and lower) at point \(P\) (distance from the straight - wire segments to \(P\) is \(R\)):
The magnetic field due to the upper straight - wire segment \(B_{1u}=\frac{\mu_0I_1}{4\pi R}\) (using the right - hand rule, direction is into the page) and due to the lower straight - wire segment \(B_{1l}=\frac{\mu_0I_1}{4\pi R}\) (using the right - hand rule, direction is out of the page). So, the net magnetic field due to the straight - wire segments \(B_{straight}=0\) (since \(B_{1u}\) and \(B_{1l}\) cancel each other).

Step2: Magnetic field due to circular arc

The magnetic field due to a circular arc of radius \(R\) carrying current \(I\) subtending an angle \(\theta\) (in radians) at the center is given by \(B=\frac{\mu_0I\theta}{4\pi R}\). For a full - circle \(\theta = 2\pi\), and for a semi - circle \(\theta=\pi\).
The current in the circular arc is \(I_2\). The magnetic field due to the circular arc at point \(P\) is \(B_{arc}=\frac{\mu_0I_2\pi}{4\pi R}=\frac{\mu_0I_2}{4R}\) (using the right - hand rule, direction is into the page if \(I_2\) is in the clockwise direction as per the figure's assumed current flow for the circular part)

Step3: Total magnetic field

The total magnetic field at point \(P\) is \(B = B_{straight}+B_{arc}\). Since \(B_{straight} = 0\), \(B=\frac{\mu_0}{4R}(I_2 - I_1)\) (assuming the direction of \(I_1\) - related fields were considered in the cancellation step. If we re - check the right - hand rule: for the circular arc (current \(I_2\)) and the straight - wire segments (current \(I_1\)), using the right - hand rule for the circular arc (if \(I_2\) is clockwise, \(B_{arc}\) is into the page) and for the straight - wire segments, if \(I_1\) is in the direction as shown, the net field is \(B=\frac{\mu_0}{4R}(I_2 - I_1)\))

Step4: When \(I_1 = I_2\)

Substitute \(I_1 = I_2\) into the formula \(B=\frac{\mu_0}{4R}(I_2 - I_1)\)

$$B=\frac{\mu_0}{4R}(I_2 - I_2)=0$$

Answer:

The magnitude of the magnetic field at point \(P\) is \(B=\frac{\mu_0}{4R}(I_2 - I_1)\). When \(I_1 = I_2\), the magnetic field \(B = 0\)