A cell is connected across two points P and Q of a uniform circular conductor, figure. Prove that the magnetic field at its centre O will be zero.

Explanation

Let `I_1`, `I_2` be the length of the two parts PRQ and PSQ of the circular conductor and `rho` be the resistance per unit length of this conductor. Let r be the radius of circular conductor.
Resistance of the portion PRQ will be
`R_1=rhol_1=rhortheta_1`
Resistance of the portion PSQ will be
`R_2=rhol_2=rhortheta_2`
Let `I_1` and `I_2` be the currents in the portion PRQ and PSQ respectively.
The potential difference across P and Q is
`V_P-V_Q=I_1R_1=I_2R_2`
or `I_1rhortheta_1=I_2rhortheta_2` or `I_1theta_1=I_2theta_2` ...(i)
According to Biot Savart's law, the magnetic field induction at the centre O due to current through portions PRQ and PSQ will be
`B_1=(mu_0)/(4pi)(I_1theta_1)/(r)` and `B_2=(mu_0)/(4pi)(I_2theta_2)/(r)`
According to Right Hand Rule, the direction of `vecB_1` and `vecB_2` are opposite to each other, along a straight line. Hence the resultant magnetic field at O is
`B=B_1-B_2` acting upwards `=(mu_0)/(4pi)(I_1theta_1)/(r)-(mu_0)/(4p)(I_2theta_2)/(r)=0 ( :' I_1theta_1=I_2theta_2)`
Hence, the resultant magnetic field at O will be zero.