To solve this problem, we will begin by understanding the formula for specific resistance \( S_1 \) in terms of the physical dimensions of the wire and demonstrate how the length of the wire affects specific resistance.
The specific resistance \( S_1 \) of a wire is given by:
\(S_1 = X\left(\frac{\pi r^2}{L}\right)\)
Here:
The formula originates from the basic relation of resistance in terms of resistivity, length, and area of cross-section:
\(R = \rho \left(\frac{L}{A}\right)\)
where:
Now, let's analyze what happens if the length of the wire is doubled. The new length, \( L' \), is:
\(L' = 2L\)
Substituting for \( L' \) in the expression for specific resistance, we get a new specific resistance \( S_1' \):
\(S_1' = X\left(\frac{\pi r^2}{L'}\right)\)
Rewriting this with \( L' = 2L \):
\(S_1' = X\left(\frac{\pi r^2}{2L}\right)\)
However, we need to reconsider as specific resistance \( S_1 \) should be independent of length since \( S_1 \) strictly corresponds to the resistivity of the material, not dependent on geometrical changes once established initially.
The specific resistance, or resistivity, essentially remains the same regardless of doubling the length because it is a material property. Thus, \( S_1' = S_1 \).
Therefore, the correct answer is that the value of the specific resistance remains \(S_1\), which matches the given correct option:
\(S_1\)
The specific resistance (or resistivity) of a material is defined as: \(Sl = X (\frac{πr^2}{L})\) ,
where X is the resistance, r is the radius, and L is the length of the wire. Specific resistance is a material property and does not change with changes in dimensions such as length or radius. Doubling the length of the wire affects the resistance X, but the specific resistance Sl remains unchanged.
The Correct answer is: \( S_1 \)
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

