Two wires of the same material having radii in the ratio `1:2`, carry currents in the ratio `4:1`. The ratio of drift velocities of electrons in them is:

Options

A. `1 :16`
B. `16:1`
C. `1:4`
D. `4:1`

Answer

To find the ratio of drift velocities of electrons in two wires of the same material, we can follow these steps:

### Step 1: Understand the relationship between current, drift velocity, and cross-sectional area
The current \( I \) in a wire is given by the formula:
\[ I = n \cdot A \cdot e \cdot v_d \]
where:
- \( n \) = number density of charge carriers (which is the same for both wires since they are of the same material),
- \( A \) = cross-sectional area of the wire,
- \( e \) = charge of an electron,
- \( v_d \) = drift velocity of the electrons.

### Step 2: Express the cross-sectional area in terms of radius
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
Let the radii of the two wires be \( r_1 \) and \( r_2 \) such that \( \frac{r_1}{r_2} = \frac{1}{2} \). Thus, we can express the areas as:
\[ A_1 = \pi r_1^2 \quad \text{and} \quad A_2 = \pi r_2^2 \]
From the ratio of the radii, we have:
\[ \frac{A_1}{A_2} = \frac{r_1^2}{r_2^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 3: Use the current ratio to find the drift velocity ratio
We are given that the currents in the two wires are in the ratio \( \frac{I_1}{I_2} = \frac{4}{1} \). Using the current formula, we can write:
\[ \frac{I_1}{I_2} = \frac{n A_1 v_{d1}}{n A_2 v_{d2}} = \frac{A_1 v_{d1}}{A_2 v_{d2}} \]
Substituting the area ratio:
\[ \frac{4}{1} = \frac{\frac{1}{4} v_{d1}}{v_{d2}} \]

### Step 4: Solve for the ratio of drift velocities
Rearranging the equation gives:
\[ 4 = \frac{1}{4} \frac{v_{d1}}{v_{d2}} \]
Multiplying both sides by \( 4 v_{d2} \):
\[ 16 v_{d2} = v_{d1} \]
Thus, the ratio of drift velocities is:
\[ \frac{v_{d1}}{v_{d2}} = 16 \]

### Conclusion
The ratio of drift velocities of electrons in the two wires is:
\[ \frac{v_{d1}}{v_{d2}} = 16 \]

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