question_answer

PQRS is a square loop made of uniform conducting wire the current enters the loop at P and leaves at S. Then the magnetic field will be:

A)  maximum at the centre of the loop

B)  zero at the centre of loop

C)   zero at all points inside the loop

D)  zero at all points outside of the loop  

Correct Answer: B

Solution :

 At point P, the current will divide in in verse ratio of resistance if arms PQRS and PS \[i=1\,{{i}_{2}}=3\]  Hence, at the centre of coil \[B=3B{{P}_{B}}\] \[=3\frac{{{\mu }_{0}}.1}{4\pi \frac{a}{2}}(sin{{45}^{o}}+sin{{45}^{o}})\] \[=3\frac{{{\mu }_{0}}.1}{4\pi a}\times 2\sqrt{2}\] ?(1) similarly due to arm PS \[=1\times \frac{{{\mu }_{0}}\times 3i}{4\pi a}\times 2\sqrt{2}\] \[=3\frac{{{\mu }_{0}}i}{4\pi a}\times 2\sqrt{2}\] ?(2) From equation (1) and (2) we observe that these are equal and opposite Therefore, the magnetic field B = 0 at the centre.