Question

A tube of mercury with resistivity 9.84 × 10-7 Ω ∙ m has a uniform electric field of 23 N/C inside the mercury. How much current is flowing in the tube, if the radius of the tube is 0.495 mm? Question 16 options:

a. 280 A
b. 18 A
c. 180 A
d. 29 A"

Answer 1

The current flowing in the tube is approximately 280 A.

Step-by-step explanation:

To calculate the current flowing in the tube, we can use Ohm's Law, which states that the current is equal to the electric field divided by the resistivity multiplied by the cross-sectional area. In this case, the cross-sectional area is the area of a circle, which can be calculated using the formula A = πr^2, where r is the radius of the tube.

Given: resistivity (ρ) = 9.84 × 10-7 Ω ∙ m, electric field (E) = 23 N/C, radius (r) = 0.495 mm = 0.495 × 10-3 m.

Using the formula for the current (I), I = E / (ρ * A),

we can calculate the current by substituting the values: I = 23 / (9.84 × 10-7 * π * (0.495 × 10-3)^2) ≈ 280 A.

Answer 2

Current, i = 18 A

Step-by-step explanation:

Given that,

Resistivity of mercury,
`\rho=9.84* 10^(-7)\ \Omega-m`

Electric field inside the mercury, E = 23 N/C

Radius of the tune, r = 0.495 mm

We need to find the current flowing in the tube. We know that the resistance in terms of length and area is given by :

`R=\rho(l)/(A)`

Using Ohm's law, we get :

V = iR, i is current, R is resistance

So,

`(V)/(i)=\rho(l)/(A)\\\\i=(VA)/(\rho l)`

Since,
`(V)/(l)=E` (electric field)

`i=(VA)/(\rho l)\\\\i=(AE)/(\rho)\\\\i=(\pi r^2E)/(\rho)\\\\i=(\pi (0.495* 10^(-3))^2* 23)/(9.84* 10^(-7))\\\\i=17.99\ A`

or

i = 18 A

So, 18 A of current is flowing in the tube.