A current I0 flows through a metallic circular loop of radius r as shown in the figure. Resistance of the segment ABC is half that of ADC. Magnitude of magnetic field at the centre O of the loop is : [figure]
- 0 I02 r
- 0 I012 r
- 0 I04 r
- 0 I02 r
Answer
(2) 0 I012 r
Solution
Let the resistance of segment ABC be R1 and that of segment ADC be R2 = 2R1.
The two semicircular segments are connected in parallel, so the current I0 divides as:
I1 = I0 ( R2R1 + R2 ) = 23I0
I2 = I0 ( R1R1 + R2 ) = 13I0
Magnetic field at the centre of a semicircle of radius r is B = 0 I4r.
Since the currents in the two semicircles flow in opposite senses, their magnetic fields at the centre oppose each other:
Bnet = B1 - B2 = 0 I14r - 0 I24r = 04r(I1 - I2)
Bnet = 04r ( 23I0 - 13I0 ) = 0 I012r
Answer: (2)