Magnetic Field

To determine the magnetic field at the mid-point of a square wire carrying a current , we use the Biot-Savart Law. A square wire consists of four straight segments. We need to calculate the contribution to the magnetic field from each segment at the center of the square and sum them up.

The Biot-Savart Law states:

Where:

• is the infinitesimal magnetic field contribution from an infinitesimal length
• is the permeability of free space
• is the current through the wire
• is the position vector from the current element to the point of interest
• is the magnitude of

The square's center is at a distance of from each vertex. The magnetic field contribution from each side of the square can be calculated independently and is directed along the perpendicular bisector of the side due to symmetry.

The length of each side of the square is , and the square is symmetrical, so the magnetic fields from opposite sides add up. Consider one side of the square:

The total magnetic field at the center from a side is:

for each side is , thus .

The contribution of magnetic field at the center is:

from one side, and the total is:

Simplifying gives the final magnitude:

The correct answer from the options provided is: .

To analyze the relationship between magnetic field components in different planes, we examine two mutually perpendicular planes inclined to the magnetic meridian.

1. Problem Setup:
Consider two planes making angles θ and (90°-θ) with the magnetic meridian. The vertical component (B_V) remains constant, while the effective horizontal components differ:

• Plane 1: B₁ = B_Hcosθ
• Plane 2: B₂ = B_Hsinθ

2. Tangent Relationships:
For each plane, we express the angle of dip:

• Plane 1: cotθ₁ = (B_Hcosθ)/B_V
• Plane 2: cotθ₂ = (B_Hsinθ)/B_V

3. Deriving the Relationship:
Squaring and adding both expressions:

cot²θ₁ + cot²θ₂ = (B_H²cos²θ)/B_V² + (B_H²sin²θ)/B_V²
= B_H²(cos²θ + sin²θ)/B_V²
= B_H²/B_V²
= cot²θ

where θ is the angle of dip in the magnetic meridian plane.

Final Result:
The relationship between the angles is given by:
cot²θ₁ + cot²θ₂ = cot²θ

The magnetic dipole moment of a current loop is given by: Where: - is the current, - is the area of the loop, and - is the unit vector normal to the plane of the loop. Here, the radius of the coil , so the area is: The magnetic dipole moment is: Since the current flows in the clockwise direction when viewed from above, the unit vector points in the direction.
Thus, the magnetic dipole moment is in the direction.

In this problem, we are asked to determine which configuration of current in square frames produces a non-zero magnetic field at the centre of the frame.
Case A: Current in opposite directions along opposite sides of the square In this case, the currents flowing along opposite sides of the square will produce magnetic fields that cancel each other out at the centre of the square. Therefore, the magnetic field at the centre is zero.
Case B: Current in the same direction along opposite sides In this case, the currents flowing in the same direction along opposite sides of the square will produce magnetic fields at the centre that also cancel out due to symmetry. Hence, the magnetic field at the centre remains zero.
Case C: Current in the same direction along all sides of the square In this configuration, all the currents flow in the same direction along the sides of the square. This produces a net magnetic field at the centre of the square, as the individual contributions from each side add up rather than cancel. Therefore, the magnetic field at the centre is non-zero.
Case D: Current in opposite directions along adjacent sides In this case, the currents along adjacent sides of the square will create opposing magnetic fields, resulting in a cancellation at the centre of the square.
Hence, the magnetic field at the centre is zero. Therefore, the correct answer is option (C) because the magnetic field at the centre of the frame is non-zero when the current flows in the same direction along all sides of the square.