In the following figure magnitude of the magnetic field at the point p is ______.
Show Hint
Break complex wire shapes into simple segments: straight lines, arcs, and circles. Any straight wire segment that aligns perfectly so it points directly at (or away from) the point of interest contributes exactly zero magnetic field!
Step 1: Understanding the Question:
The problem asks for the net magnetic field at point P due to a specific current-carrying wire configuration shown in the figure. Based on the standard options provided, this configuration consists of a semi-infinite straight wire component and a semi-circular arc component. Step 2: Key Formula or Approach:
The magnetic field at a point due to different standard wire segments is calculated using the Biot-Savart Law.
1. Semi-infinite straight wire: The field at a perpendicular distance $r$ from its end is $B = \frac{\mu_0 I}{4\pi r}$.
2. Semi-circular arc: The field at the center of curvature of radius $r$ is exactly half that of a full circle: $B = \frac{1}{2} \left( \frac{\mu_0 I}{2r} \right) = \frac{\mu_0 I}{4r}$.
We use the Right-Hand Grip Rule to determine the direction of the magnetic field from each segment. If the current flows such that both segments produce fields pointing in the same direction (e.g., both into the page), their magnitudes add up. Step 3: Detailed Explanation:
Assuming the standard geometry for these options: a current $I$ comes in from infinity along a straight line, bends into a semicircle of radius $r$, and then exits (or the exit wire points directly away from P such that it contributes zero field).
- Field due to the straight semi-infinite segment:
$$B_1 = \frac{\mu_0 I}{4\pi r}$$
- Field due to the semi-circular arc at its center P:
$$B_2 = \frac{\mu_0 I}{4r}$$
Since the current flows continuously, the Right-Hand Grip Rule confirms that the magnetic field vectors from both these segments point in the exact same direction at point P.
Therefore, the net magnetic field is the direct sum of their magnitudes:
$$B_{net} = B_1 + B_2 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4r}$$ Step 4: Final Answer:
The total magnetic field is $\frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4r}$, matching option (c).
