A current $I$ flows through an infinitely long conductor $A B C$ bent to form a right angle. (Fig).
How many times will the intensity of the magnetic field change at point $M$ if an infinitely long straight conductor $B D$ is so connected to point $B$ that the current $I$ branches at point $B$ into two equal parts and the current in the conductor $A B$ remains the same ?

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$\frac{H_{2}}{H_{1}} = \frac{5}{2}$

$\frac{H_{2}}{H_{1}} = \frac{3}{2}$

$\frac{H_{2}}{H_{1}} = 2$

$\frac{H_{2}}{H_{1}} = \frac{2}{3}$

answer is D.

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Detailed Solution

The conductor $B C$ does not induce a field at point $M$ lying on the continuation of $B C$ .
Since the intensity of the field is proportional to the current, then $H_{1} = k I$ without conductor $B D$ . The fields from the conductors $A B$ and $B D$ are added. Therefore after conductor $B D$ is connected
$H_{2} = k I + \frac{k I}{2}$
whence $\frac{H_{2}}{H_{1}} = \frac{3}{2}$ .

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