Question

A square loop of side 'a' is made by a current carrying wire. Magnetic field at its vertex 'P' is

• A. $\frac{{\mu_0 I}}{{2\pi a}}$
• B. $\frac{{\mu_0 I}}{{4\pi a}}$
• C. $\frac{{\mu_0 I}}{{a}}$
• D. $0$

Answer 1

Answer: (D)

0

Answer 2

The magnetic field at a vertex of a square loop carrying current is the vector sum of the magnetic fields produced by the two sides of the square that do not pass through that vertex. Let the square loop have side length 'a' and carry a current 'I'. Let P be one of the vertices.

Consider a square loop ABCD with vertices A=(0,0), B=(a,0), C=(a,a), and D=(0,a). Let P be vertex A. Assume the current flows counter-clockwise (A→B→C→D→A).

The magnetic field at P due to the sides AB and AD (which pass through P) is considered infinite if P is exactly on the wire. However, in standard problems of this type, the question implies calculating the field due to the other two sides of the loop, or it refers to a point infinitesimally close to the vertex but not on the wire itself. Assuming the latter, and that the field contributions from the adjacent sides are either not considered or cancel out in some limiting sense, we calculate the field from the non-adjacent sides.

The sides contributing to the magnetic field at vertex A are BC and CD.

1. Magnetic field due to side BC:

   • Side BC runs from B(a,0) to C(a,a). The current flows in the +y direction.
   • The point P is at A(0,0).
   • The perpendicular distance from P to the line containing BC (which is the line x=a) is $r = a$.
   • The magnetic field due to a finite straight wire segment is given by $B = \frac{{\mu_0 I}}{{4 \pi r}} (\sin \theta_1 + \sin \theta_2)$, where $\theta_1$ and $\theta_2$ are the angles subtended by the ends of the wire segment at the point, measured from the perpendicular.
   • For side BC, the perpendicular from P to the line x=a lies along the x-axis.
   • The line from P to B(a,0) is along the x-axis. The angle it makes with the perpendicular (x-axis) is $\theta_1 = 0^\circ$.
   • The line from P to C(a,a) makes an angle of $45^\circ$ with the x-axis. So, $\theta_2 = 45^\circ$.
   • $B_{BC} = \frac{{\mu_0 I}}{{4 \pi a}} (\sin 0^\circ + \sin 45^\circ) = \frac{{\mu_0 I}}{{4 \pi a}} (0 + \frac{1}{{\sqrt{2}}}) = \frac{{\mu_0 I}}{{4 \pi a \sqrt{2}}}$.
   • Using the right-hand rule (thumb along current, fingers curl), the magnetic field at P due to BC is directed out of the page (+z direction).

2. Magnetic field due to side CD:

   • Side CD runs from C(a,a) to D(0,a). The current flows in the -x direction.
   • The point P is at A(0,0).
   • The perpendicular distance from P to the line containing CD (which is the line y=a) is $r = a$.
   • The perpendicular from P to the line y=a lies along the y-axis.
   • The line from P to C(a,a) makes an angle of $45^\circ$ with the y-axis. So, $\theta_1 = 45^\circ$.
   • The line from P to D(0,a) is along the y-axis. The angle it makes with the perpendicular (y-axis) is $\theta_2 = 0^\circ$.
   • $B_{CD} = \frac{{\mu_0 I}}{{4 \pi a}} (\sin 45^\circ + \sin 0^\circ) = \frac{{\mu_0 I}}{{4 \pi a}} (\frac{1}{{\sqrt{2}}} + 0) = \frac{{\mu_0 I}}{{4 \pi a \sqrt{2}}}$.
   • Using the right-hand rule (thumb along current, fingers curl), the magnetic field at P due to CD is directed into the page (-z direction).

The total magnetic field at vertex P is the vector sum of $B_{BC}$ and $B_{CD}$. Since they are equal in magnitude and opposite in direction, they cancel each other out: $B_{net} = B_{BC} + B_{CD} = \frac{{\mu_0 I}}{{4 \pi a \sqrt{2}}} (\hat{k}) + \frac{{\mu_0 I}}{{4 \pi a \sqrt{2}}} (-\hat{k}) = 0$.

The same result is obtained if the current direction is reversed.