Question

A wire carrying current I and other parallel wire carrying current 2I in the same direction produces a magnetic field B at the midpoint between them. Then the magnitude of field at the same point, when the 2I wire is switched off

• A. B/2
• B. 2 B
• C. B
• D. 4 B

Hint:

Parallel currents in the same direction subtract fields at the midpoint. Since $2I - I = I$, turning off the $2I$ wire leaves the field of $I$ alone, keeping the magnitude identical!

Answer 1

The Correct Option is C

Step 1: Field of a straight wire.
A long wire makes a field $B = \dfrac{\mu_{0}I}{2\pi r}$ at distance $r$. The direction is set by the right-hand rule.

Step 2: Set the midpoint distance.
Let the wires be $2d$ apart. The midpoint is $d$ from each wire.

Step 3: Fields at the midpoint.
Since the currents are in the same direction, at the midpoint their fields point opposite ways. So they subtract.

Step 4: Write each field.
Wire 1 (current $I$): $B_{1} = \dfrac{\mu_{0}I}{2\pi d}$. Wire 2 (current $2I$): $B_{2} = \dfrac{\mu_{0}(2I)}{2\pi d} = 2B_{1}$.

Step 5: Net field with both on.
\[ B = B_{2} - B_{1} = 2B_{1} - B_{1} = B_{1} \]So the given net field $B$ equals $B_{1}$.

Step 6: Switch off the $2I$ wire.
Now only wire 1 is left, giving field $B_{1}$. Since $B = B_{1}$, the field is still $B$. This is option 3.
\[ \boxed{\text{Field} = B} \]