Figure shows a square loop ABCD with edge length `a`. The resistance of the wire ABC is `r` and that of ADC is `2r`. Find the magnetic field B at the centre of the loop assuming uniform wires. `

Options

A. `(sqrt2 mu_(0) i)/(3pi a) "" odot`
B. `(sqrt2 mu_(0)i)/(3 pi a) "" otimes`
C. `(sqrt2 mu_(0)i)/(pi a) "" odot`
D. `(sqrt2 mu_(0) i)/(pi a) "" otimes `

Answer

B

Explanation

According to question, resistance of wire ADC is twice that of wire ABC. Hence current flowing through ADC is half that of ABC i.e., `(i_(2))/(i_(1)) = (1)/(2)` Also `i_(1) + i_(2) = i`
`implies i_(1) = (2i)/(3)` and `i_(2) = (i)/(3)`
Magnetic field at centre O due to wire AB and BC (part 1 and 2)
`B_(1) = B_(2) = (mu_(0))/(4pi) (2i_(1) sin 45^(@))/(a //2) = (mu_(0))/(4pi) * (2 sqrt2 i_(1))/(a)`
and magnetic field at centre O due to wire AD and DC [i.e part 3 and 4] `B_(3) = B_(4) = (mu_(0))/(4pi) (2sqrt2 i_(2))/(a)`
Also `i_(1) = 2i_(2)` . So (`B_(1) = B_(2) ) gt (B_(3) = B_(4))`
Hence net magnetic field at centre O
`B_("net") = (B_(1) + B_(2)) - (B_(3) + B_(4))`
`= 2xx (mu_(0))/(4pi) * (2 sqrt2 xx ((2)/(3) i))/(a) - (mu_(0))/(4pi ) * (2 sqrt2 ((i)/(3)) xx 2)/(a)`
`= (mu_(0))/(4pi) * (4sqrt2i)/(3a) (2- 1) = (sqrt2 mu_(0) i)/(3pi a)`