Question

Two wires \(A\) and \(B\) of the same material having lengths \(L_A, L_B\) and radii \(R_A, R_B\) and drift velocities \(V_A, V_B\) respectively carry the same current. If \(L_A=L_B\) and \(R_A=2R_B\), then the value of \[ \left(\frac{V_A}{V_B}\right) \] is:

• A. \(0.25\)
• B. \(0.5\)
• C. \(2.0\)
• D. \(1.0\)

Hint:

For wires carrying the same current and made of the same material, \[ I=neAv_d \] implies \[ v_d \propto \frac{1}{A} \] Thus, drift velocity is inversely proportional to the cross-sectional area of the wire.

Answer 1

The Correct Option is A

Step 1: Use the drift velocity relation.
Current \( I = neAv_d \). For the same current and same material (same n, e): \( A_A V_A = A_B V_B \).

Step 2: Substitute the radius condition \( R_A = 2R_B \).
\( A_A = \pi R_A^2 = 4\pi R_B^2 = 4A_B \). Therefore: \( 4A_B V_A = A_B V_B \Rightarrow \frac{V_A}{V_B} = \frac{1}{4} \) \[ \boxed{0.25} \]