Figure shows a current carrying square loop ABCD of edge length 'a' lying in a plane. If the resistance of the ABC part is r and that of ADC part is 2r, then the magnitude of the resultant magnetic field at centre of the square loop is [figure]
- 2 0 I3 a
- 2 0 I3 a
- 3 0 I 2a
- 0 I2 a
Answer
(2) 2 0 I3 a
Solution
ABC (resistance r) and ADC (resistance 2r) are in parallel:
i1 = 2rr + 2rI = 2I3, i2 = I3
Field at the centre due to one side (distance a2, each end subtending 45 ):
Bside = 0 i4 ( a2 )( 45 + 45 ) = 2 0 i2 a
Path ABC is clockwise and path ADC is anticlockwise about the centre, so their fields are opposite.
B = 2 2 02 a(i1 - i2) = 2 0 a I3
B = 2 0 I3 a
Answer: (2)