Question

In the rainy season, the amazon flows fast and runs deep. In one location, the river is 23m deep and moves at a speed of 4.0m/s toward the east. The earth's 50uT magnetic field is parallel to the ground and directed northward. If the bottom of the river is at 0V, what is the potential (magnitude and sign) at the surface?

Answer 1

Concepts and reason

The concept required to solve this problem is potential energy. Initially, write an expression for the potential energy. Later, calculate the magnitude of the generated potential energy. Finally, using the right-hand thumb rule find the direction.

Fundamentals

The expression for the potential energy generated is as follows:

\(E=B l v\)

Here, \(B\) is the magnetic field, \(l\) is the length, and \(v\) is the velocity. According to the right-hand thumb rule, grab the wire with your right hand in such a way that the thumb points in the direction of current and the curled finger tell about the direction of the magnetic field.

 

Substitute \(4.0 \mathrm{~m} / \mathrm{s}\) for \(\mathrm{v}, 23 \mathrm{~m}\) for \(\mathrm{I}\), and \(50 \mu \mathrm{T}\) for \(\mathrm{B}\) in the equation \(E=B l v\)

$$ \begin{array}{c} E=(50 \mu \mathrm{T})(23 \mathrm{~m})(4.0 \mathrm{~m} / \mathrm{s}) \\ =(50 \mu \mathrm{T})\left(\frac{10^{-6} \mathrm{~T}}{1.0 \mu \mathrm{T}}\right)(23 \mathrm{~m})(4.0 \mathrm{~m} / \mathrm{s}) \\ =4.6 \times 10^{-3} \mathrm{~V} \\ =4.3 \mathrm{mV} \end{array} $$

The magnetic field is given in micro Tesla to convert it into Tesla it is multiplied by a factor of \(10^{-6}\). Also, the magnitude of the energy is \(4.3 \mathrm{mV}\), because, the term \(10^{-3}\) represents milli.

 

Using the right-hand thumb rule, the induced current is directed towards the top. Thus, the sign of the potential is positive.

The magnitude of the potential energy is \(4.6 \mathrm{mV}\) and the sign is positive.

The potential at the upper surface is positive, because, the positive charge accumulates on the upper surface due to the magnetic force.