An infifnitely long conductor `PQR` is bent to form a right angle as shown in figure. A current `I` flows through `PQR`. The magnetic field due to this current at the point `M` is `H_1` Now, another infinitely long straight conductor `QS` is connected at `Q`, so that current is `I/2` in `QR` as well as in `QS`, the current in `PQ` remaining unchanged. The magnetic field at `M` is now `H_2`. The ratio `H_1/H_2` is given by `

Options

A. `1//2`
B. `1`
C. `2//3`
D. `2`

Answer

C

Explanation

(c) `vecB_1=(mu_0I)/(4piR)odot, vecB_2=(mu_0I)/(4piR)+(mu_0I//2)/(4piR)odot`
`implies B_2=(mu_0I)/(4piR)(3/2)odot implies |(B_1)/(B_2)|=2/3`