Question

A square loop is made by a uniform conductor wire as shown in figure. Find the net magnetic field at the centre of the loop if the side length of the square is a.

A.$\dfrac{{{\mu }_{0}}i}{2a}$
B.Zero
C.$\dfrac{{{\mu }_{0}}{{i}^{2}}}{{{a}^{2}}}$
D.None of these

Answer 1

Hint: To find the solution for this question first, find the current through each wire of the square. Find the magnetic field at the centre due to each wire. Then find the direction of the magnetic field produced for each side and add these four to get the final solution. Since, the magnitude of the field produced will be the same, we can find the answer directly on the basis of direction of magnetic field.

Complete answer:
As shown in the picture current I is flowing from the battery and at the loop at point p it is divided. Because this is a square loop and made of the same material the resistance of the wire in both sides will be the same. So, equal current will pass through the wires. Along PS the current will be $I/2$ and along PQ the current will be $I/2$.
The magnitude of the magnetic field produced at the centre of the square loop due to the current flowing through all four sides will be the same because the length of the wires or the distance of the centre from all the four sides are equal.
Let this magnetic field be B.
Now, direction of the magnetic field due to a current element can be given by the right-hand thumb rule. The right-hand thumb rule states that if we point the thumb of our right hand in the direction of the current through the wire, the fingers will curl up in the direction of the magnetic field.
Now consider the current through the side wire PS and PR. Current through both the wire is same and direction of current is also same. If we apply the right-hand thumb rule, we will get the direction of the magnetic field as going inside the plane of the square.
Now If we consider the flow of current through the wires PQ and QS the direction of the magnetic field will be out of the plane of the paper.
So, we get 2B magnetic field out of paper and 2B magnetic field inside the paper or -2B magnetic field outside the paper.
So, the total magnetic field will be the sum of these two, which will be zero.
The correct option is (B)

Note: The magnetic field direction and the magnetic field magnitude around a current carrying wire can be given by the ampere’s law and Biot-savart law. The direction is simply given by right-hand thumb rule. It can also give the direction of current in a wire placed in a magnetic field.