A current I flows in an infinitely long wire with cross-section in the form of a semi-circular ring

Q: Why do we use the formula for a straight wire and not a circular loop?

Because the current is flowing *along* the length of the infinitely long wire, not in a circular loop. The cross-section is semi-circular, but the current path is straight.

Q: How did we get $dI = \frac{I}{\pi} d\theta$?

The total current $I$ is uniformly distributed over the entire semi-circular arc, which corresponds to an angle of $\pi$ radians. So, the current per unit angle is $I/\pi$. For a small angle $d\theta$, the current is $(I/\pi) d\theta$.

Q: Why do the vertical components cancel out?

For every elemental wire at an angle $\theta$ on the right side, there is an identical elemental wire at an angle $-\theta$ on the left side. Their vertical magnetic field components point in opposite directions and have the same magnitude, thus cancelling each other out.

Q: Why are the integration limits from $-\pi/2$ to $\pi/2$?

We chose our vertical axis (y-axis) as the axis of symmetry ($\theta = 0$). To cover the entire semi-circle, we must sweep from the far left ($-\pi/2$) to the far right ($+\pi/2$).

Q: What if the current was not uniformly distributed?

If the current density was a function of $\theta$, say $J(\theta)$, then $dI$ would be $J(\theta) R d\theta$, and we would have to include this function inside our integral. The symmetry argument might also fail depending on the function.

Question

A current $I$ flows in an infinitely long wire with cross-section in the form of a semi-circular ring of radius $R$. The magnitude of the magnetic induction along its axis is

Options:

(A) $\frac{\mu_0 I}{2\pi^2 R}$

(B) $\frac{\mu_0 I}{2\pi R}$

(C) $\frac{\mu_0 I}{4\pi R}$

(D) $\frac{\mu_0 I}{\pi^2 R}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup
Imagine looking straight down the barrel of an infinitely long wire. But this isn't a normal wire; its cross-section is shaped like a semi-circular ring of radius $R$. The total current $I$ is flowing uniformly through this entire shape, let's say, into the screen.

To find the magnetic field at the center, we can't just use a single formula. We need to break this complex shape into tiny, manageable pieces. We can think of this semi-circular ring as being made up of many infinitely long, thin straight wires placed side by side.

### The Power of Superposition
Let's focus on one such tiny elemental wire. We choose an element that subtends a small angle $d	heta$ at the center, located at an angle $	heta$ from the vertical axis. Now, how much current flows through just this tiny piece? Since the total current $I$ is spread over a total angle of $\pi$, the current $dI$ in our element will be $I$ divided by $\pi$, times $d	heta$.

$$
dI = \frac{I}{\pi} d	heta
$$

Now, what is the magnetic field produced by this single, thin straight wire at the center? We know the formula for a long straight wire is $\mu_0$ times the current, divided by $2\pi r$. Here, the distance is $R$, and the current is $dI$. Substituting our value for $dI$, we get the magnitude of the small magnetic field, $dB$.

$$
dB = \frac{\mu_0 dI}{2\pi R} = \frac{\mu_0}{2\pi R} \left(\frac{I}{\pi} d	hetaight) = \frac{\mu_0 I}{2\pi^2 R} d	heta
$$

### Symmetry to the Rescue
Look closely at the direction of this magnetic field $dB$. Using the right-hand grip rule, it points perpendicular to the radius vector. We can resolve this $dB$ into two components: a horizontal component $dB \cos	heta$, and a vertical component $dB \sin	heta$.

Notice the symmetry! For every element on the right, there's a mirror element on the left. Their vertical sine components will perfectly cancel each other out. Since the vertical components vanish, the net magnetic field is simply the sum of all the horizontal components. So, we need to integrate $dB \cos	heta$.

### The Final Integration
Our semi-circle spans from the far left to the far right, which means our angle $	heta$ goes from $-\pi/2$ to $+\pi/2$. Let's execute the integration. We pull out the constants.

$$
B = \int_{-\pi/2}^{\pi/2} dB \cos	heta = \int_{-\pi/2}^{\pi/2} \frac{\mu_0 I}{2\pi^2 R} \cos	heta d	heta
$$

The integral of $\cos	heta$ is $\sin	heta$. Applying the upper limit $\pi/2$ and the lower limit $-\pi/2$, we get $\sin(\pi/2) - \sin(-\pi/2)$. That's $1 - (-1)$, which equals $2$.

$$
B = \frac{\mu_0 I}{2\pi^2 R} [\sin	heta]_{-\pi/2}^{\pi/2} = \frac{\mu_0 I}{2\pi^2 R} (2) = \frac{\mu_0 I}{\pi^2 R}
$$

Multiplying this $2$ with our constants, the $2$ in the denominator cancels out, leaving us with our final elegant answer: **$\frac{\mu_0 I}{\pi^2 R}$**. This is a classic JEE problem that beautifully combines Ampere's law concepts with calculus and symmetry.

Animated Solution for Physics - Magnetic Effects of Current: A current $I$ flows in an infinitely long wire with cross-section in the form of a semi-circular ring of radius $R$ . The magnitude of the magnetic induction along its axis is

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Elemental Wire Approach}

\text{Current in Elemental Wire}

\text{Magnetic Field } dB

\text{Resolving Components}

\text{Net Magnetic Field Integration}

\text{Final Calculation}

\text{Conclusion & Symmetry}

The Sigma Insight: Ampere's Circuital Law

Visualizing the Setup Imagine looking straight down the barrel of an infinitely long wire

The Power of Superposition Let's focus on one such tiny elemental wire

Symmetry to the Rescue Look closely at the direction of this magnetic field $d B$

The Final Integration Our semi-circle spans from the far left to the far right, which means our angle $θ$ goes from $- π /2$ to $+ π /2$

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