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:
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:
Show 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.
Updated On: Jun 26, 2026
- \(0.25\)
- \(0.5\)
- \(2.0\)
- \(1.0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the relation between current and drift velocity.
Current in a conductor is given by \[ I=neAv_d \] where \[ n=\text{number density of charge carriers} \] \[ e=\text{charge of electron} \] \[ A=\text{cross-sectional area} \] \[ v_d=\text{drift velocity} \]
Step 2: Use the condition of same material and same current.
Since both wires are made of the same material, \[ n=\text{constant} \] and \[ e=\text{constant} \] Also, both wires carry the same current. Therefore, \[ neA_AV_A=neA_BV_B \] \[ A_AV_A=A_BV_B \] Hence, \[ \frac{V_A}{V_B} = \frac{A_B}{A_A} \]
Step 3: Express area in terms of radius.
The cross-sectional area of a wire is \[ A=\pi R^2 \] Thus, \[ \frac{V_A}{V_B} = \frac{\pi R_B^2}{\pi R_A^2} = \frac{R_B^2}{R_A^2} \] Given, \[ R_A=2R_B \] Substituting, \[ \frac{V_A}{V_B} = \frac{R_B^2}{(2R_B)^2} \] \[ \frac{V_A}{V_B} = \frac{R_B^2}{4R_B^2} \] \[ \frac{V_A}{V_B} = \frac{1}{4} \] \[ \frac{V_A}{V_B}=0.25 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{V_A}{V_B}=0.25} \]
Current in a conductor is given by \[ I=neAv_d \] where \[ n=\text{number density of charge carriers} \] \[ e=\text{charge of electron} \] \[ A=\text{cross-sectional area} \] \[ v_d=\text{drift velocity} \]
Step 2: Use the condition of same material and same current.
Since both wires are made of the same material, \[ n=\text{constant} \] and \[ e=\text{constant} \] Also, both wires carry the same current. Therefore, \[ neA_AV_A=neA_BV_B \] \[ A_AV_A=A_BV_B \] Hence, \[ \frac{V_A}{V_B} = \frac{A_B}{A_A} \]
Step 3: Express area in terms of radius.
The cross-sectional area of a wire is \[ A=\pi R^2 \] Thus, \[ \frac{V_A}{V_B} = \frac{\pi R_B^2}{\pi R_A^2} = \frac{R_B^2}{R_A^2} \] Given, \[ R_A=2R_B \] Substituting, \[ \frac{V_A}{V_B} = \frac{R_B^2}{(2R_B)^2} \] \[ \frac{V_A}{V_B} = \frac{R_B^2}{4R_B^2} \] \[ \frac{V_A}{V_B} = \frac{1}{4} \] \[ \frac{V_A}{V_B}=0.25 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{V_A}{V_B}=0.25} \]
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