Question

A straight wire of diameter $0.4 \text{ mm}$ carrying a current of $2 \text{ A}$ is replaced by another wire of $0.8 \text{ mm}$ diameter carrying the same current. The magnetic field at distance (R) from both the wires is '$B_1$' and '$B_2$' respectively. The relation between $B_1$ and $B_2$ is

• A. $B_1 = \frac{B_2}{2}$
• B. $B_1 = B_2$
• C. $B_1 = 2 B_2$
• D. $B_1 = \frac{B_2}{3}$

Hint:

Don't fall into the trap of using wire dimensions for external fields! A wire's radius or diameter changes the internal magnetic field gradient ($\text{inside } B \propto r$), but for any point outside the wire, the system functions exactly like an infinitely thin line current.

Answer 1

The Correct Option is B

Step 1: Compare the two wires.
The first wire has diameter $0.4\ \text{mm}$, the second $0.8\ \text{mm}$. Both carry the same $2\ \text{A}$. We compare the fields $B_1$ and $B_2$ at the same distance $R$ from each.
Step 2: Field of a long straight wire.
Outside a long straight conductor, $B = \dfrac{\mu_0 I}{2\pi R}$.
Step 3: See which quantities matter.
The formula contains only the current $I$ and the distance $R$. The diameter of the wire does not appear.
Step 4: Check the point lies outside.
Since $R$ is measured from the wire and the point is external, the simple formula applies for both wires.
Step 5: Compare the two cases.
Both wires have the same current ($2\ \text{A}$) and the field is taken at the same $R$, so $B_1 = \dfrac{\mu_0(2)}{2\pi R}$ and $B_2 = \dfrac{\mu_0(2)}{2\pi R}$ are identical.
Step 6: Conclude.
The different diameters change nothing externally, so $B_1 = B_2$.
\[ \boxed{B_1 = B_2\ \text{(option 2)}} \]