Question 616: Current Electricity

Question

A uniform wire of diameter $$d$$ carries a current of $$100 \mathrm{~mA}$$ when the mean drift velocity of electrons in the wire is $$v$$. For a wire of diameter $$\frac{d}{2}$$ of the same material to carry a current of $$200 \mathrm{~mA}$$, the mean drift velocity of electrons in the wire is

Accepted Answer

To solve this problem, we need to understand the relationship between the current, the drift velocity, and the cross-sectional area of the wire. The electric current $$I$$ in a wire is given by
$$I = n e A v_d$$
where:
$$n$$ is the number density of electrons,
$$e$$ is the charge of an electron,
$$A$$ is the cross-sectional area of the wire, and
$$v_d$$ is the mean drift velocity of the electrons.
Let's denote the current in the thicker wire as $$I_1 = 100 \mathrm{~mA} = 0.1 \mathrm{~A}$$ and the current in the thinner wire as $$I_2 = 200 \mathrm{~mA} = 0.2 \mathrm{~A}$$.
The diameter of the thicker wire is $$d$$, so its cross-sectional area, $$A_1$$, is
$$A_1 = \pi \left(\frac{d}{2}ight)^2 = \frac{\pi d^2}{4}$$
For the thinner wire, the diameter is $$\frac{d}{2}$$, so its cross-sectional area, $$A_2$$, is
$$A_2 = \pi \left(\frac{d}{4}ight)^2 = \frac{\pi d^2}{16}$$
Using the relation for current in the thicker wire, we have
$$I_1 = n e A_1 v = n e \left(\frac{\pi d^2}{4}ight) v = 0.1 \mathrm{~A}$$
Let the mean drift velocity in the thinner wire be denoted by $$v'$$. For the thinner wire:
$$I_2 = n e A_2 v' = n e \left(\frac{\pi d^2}{16}ight) v' = 0.2 \mathrm{~A}$$
Now, we can equate the expressions for current and solve for $$v'$$:
$$\frac{\pi d^2}{16} n e v' = 0.2 \mathrm{~A}$$
$$v' = \frac{0.2 \mathrm{~A}}{n e \left(\frac{\pi d^2}{16}ight)}$$
From the equation for the thicker wire:
$$0.1 \mathrm{~A} = \frac{\pi d^2}{4} n e v$$
$$v = \frac{0.1 \mathrm{~A}}{n e \left(\frac{\pi d^2}{4}ight)}$$
Dividing the two equations, we get
$$\frac{v'}{v} = \frac{0.2 \mathrm{~A}}{\left(\frac{\pi d^2}{16}ight) n e} 	imes \frac{(\frac{\pi d^2}{4}) n e}{0.1 \mathrm{~A}} = \frac{0.2 \mathrm{~A} 	imes 4}{0.1 \mathrm{~A} 	imes 16} = \frac{4}{1} = 8$$
Thus,
$$v' = 8 v$$
The mean drift velocity of electrons in the thinner wire is 8 times the mean drift velocity in the thicker wire. Therefore, the correct option is:
Option B
: $$8 v$$

Answer

$$4 v$$

Answer

$$8 v$$

Answer

$$v$$

Answer

$$2 v$$

Current Electricity

A uniform wire of diameter $d carries a current of 100 \mathrm{~mA} when the mean drift velocity of electrons in the wire is v. For a wire of diameter \frac{d}{2} of the same material to carry a current of 200 \mathrm{~mA}$, the mean drift velocity of electrons in the wire is

Choose the correct answer: