Question

7) A wire with resistance of
8.0Ω
is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire assuming that the resistivity and density of the material are unaffected by the drawing process.

a) 72Ω

b) 60Ω

c) 80Ω

d) 45Ω

8) A cirrent flows in a wire of circular cross-section with the free electrons travelling with a mean drift velocity v. If an equal current flows in a wire of the same material but of twice theradius, what is the new mean drift velocity?
a) v/4
b) v/2
c) 2v
d) 4v

Answer 1

Answer on question #61339, Physics / Electromagnetism

7) A wire with resistance of $8.0\Omega$ is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire assuming that the resistivity and density of the material are unaffected by the drawing process.

a) $72\Omega$

b) $60\Omega$

c) $80\Omega$

d) $45\Omega$

Solution:

The electrical resistivity $\rho$ is defined as:

$$\rho = R \frac {A}{l}$$

Thus, the resistance is

$$R = \rho \frac {l}{A}$$

The volume of wire is

$$V = A l = c o n s t$$ $$A _ {1} l _ {1} = A _ {2} l _ {2}$$ $$\frac {l _ {1}}{l _ {2}} = \frac {A _ {2}}{A _ {1}}$$

By the condition of the task

$$l _ {2} = 3 l _ {1}$$

The ratio of resistances is

$$\frac {R _ {1}}{R _ {2}} = \frac {\rho l _ {1} A _ {2}}{\rho l _ {2} A _ {1}}$$

So,

$$\frac {R _ {1}}{R _ {2}} = \frac {l _ {1}}{3 l _ {1}} \cdot \frac {l _ {1}}{3 l _ {1}}$$ $$\frac {R _ {1}}{R _ {2}} = \frac {1}{3} \cdot \frac {1}{3}$$ $$\frac {R _ {1}}{R _ {2}} = \frac {1}{9}$$ $$R _ {2} = 9 R _ {1} = 9 \cdot 8.0 \Omega = 72 \Omega$$

Answer: a) 72Ω

8) A current flows in a wire of circular cross-section with the free electrons travelling with a mean drift velocity $v$. If an equal current flows in a wire of the same material but of twice the radius, what is the new mean drift velocity?

a) $v/4$

b) $v/2$

c) $2v$

d) $4v$

**Solution:**

$$j = nev$$ $$j = \frac{l}{A}$$

Where,

$$A = \pi r^2$$ $$j = \frac{l}{\pi r^2}$$ $$\frac{l}{\pi r^2} = nev$$

For wire twice the radius,

$$\frac{l}{\pi r_1^2} = nev_1$$

By the condition of the task,

$$r_1 = 2r$$ $$\frac{l}{\pi (2r)^2} = nev_1$$ $$\frac{l}{4\pi r^2} = nev_1$$ $$\frac{nev}{4} = nev_1$$

The new mean drift velocity is

$$v_1 = \frac{v}{4}$$

**Answer:** a) $v/4$

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