Question

An infinitely 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

• A. 44563
• B. 1
• C. 44595
• D. 2

Answer 1

Answer: (C)

44595

Answer 2

$H_1$ = Magnetic field at M due to PQ + Magnetic field at M due to QR
But magnetic field at M due to QR = 0
$\therefore$ Magnetic field at M due to PQ (or due to current 1 in PQ)$=H_1$
Now $H_2$ = Magnetic field at M due to PQ (current I)
$\, \, \, \, \, \, \, \, \, \,$+ magnetic field at M due to QS (current 1/2)
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$+ magnetic field at M due to QR
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =H_1+\frac{H_1}{2}+0=\frac{3}{2}H_1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{H_1}{H_2}=\frac{2}{3}$
NOTE Magnetic field at any point lying on the current carrying straight conductor is zero.