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

As per question, it is clear that curent through QS should be `I//2` in the direction Q to S.
Magnetic field induction at a distance r from a long straight conductor carrying current I is
`B=(mu_0)/(4pi)I/r(sinphi_1+sinphi_2)`
As the point M is at the end side of a long straight conductor, so `phi_1=90^@` and `phi_2=0^@`. Therefore magnetic field induction at M due to current through PQ is
`B=(mu_0)/(4pi)I/r(sin 90^@+sin 0^@)`
`=(mu_0)/(4pi)I/r`, acting perpendicular to the plane of paper outwards.
The magnetic field induction at M due to current in QR will be zero. As per question
`H_1=(mu_0)/(4pi)I/r`
When straight conductor QS is connected at Q,
then `H_2=(mu_0)/(4pi)I/r+(mu_0)/(4pi)(I//2)/(r)=(mu_0)/(4pi)(3/2I/r)`
So `(H_1)/(H_2)=(mu_0I//(4pir))/(mu_0/4pi(3/2I/r))=2/3`