Question

19 A nichrome wire is 1.0m long and
1.0mm2
in cross-sectional area. It carries a current of 4.0 A when a potential difference of 2 V is applied between its ends. Calculate the conductivity of the wire.
(2MΩm)−1
(4kΩm)−1
(2mΩm)−1
(4Ωm)−1

20 The current I in a conductor as a function of time t is given as
I(t)=5t2−3t+10
where current is in ampres A and t is in seconds s. What quantity of charge moves across a section through the conductor during the interval t=2s to t=5s?
154.4C
193.5C
225.5C
300.0C

Answer 1

Answer on Question #55897, Physics / Electromagnetism

19 A nichrome wire is $1.0 \, \text{m}$ long and $1.0 \, \text{mm}^2$ in cross-sectional area. It carries a current of 4.0 A when a potential difference of $2 \, \text{V}$ is applied between its ends. Calculate the conductivity of the wire.

Solution:

The conductivity of a wire can be expressed as

$$\sigma = \frac{L}{RA}$$

where

- $L$ = length = 1.0 m

- $A$ = cross sectional area = $1.0 \times 10^{-6} \, \text{m}^2$.

The resistance is

$$R = \frac{V}{i} = \frac{2 \, \text{V}}{4.0 \, A} = 0.5 \, \Omega$$

Thus,

$$\sigma = \frac{1.0}{0.5 \cdot 1.0 \cdot 10^{-6}} = 2 \cdot 10^6 \, (\Omega \, \text{m})^{-1}$$

Answer: $(2 \, \text{M} \, \Omega \, \text{m})^{-1}$

20 The current I in a conductor as a function of time t is given as

$$I(t) = 5t^2 - 3t + 10$$

where current is in ampres A and t is in seconds s. What quantity of charge moves across a section through the conductor during the interval $t = 2 \, \text{s}$ to $t = 5 \, \text{s}$?

Solution:

The current I is the time rate of transfer of charge across a cross section, so here we have

$$q = \int_{t_1}^{t_2} I(t) \, dt$$

Thus,

$$\begin{aligned} q &= \int_{2}^{5} (5t^2 - 3t + 10) \, dt = \left. \left(\frac{5}{3} t^3 - \frac{3}{2} t^2 + 10t\right) \right|_{2}^{5} \\ &= \frac{5 \cdot 5^3}{3} - \frac{3 \cdot 5^2}{2} + 10 \cdot 5 - \left( \frac{5 \cdot 2^3}{3} - \frac{3 \cdot 2^2}{2} + 10 \cdot 2 \right) = 193.5 \, \text{C} \end{aligned}$$

Answer: 193.5 C

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