Question

What force is exerted on the water in an MHD drive utilizing a${}^{\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{‐}\mathbf{cm}}$-diameter tube, iflocalid="1654594579558" ${}^{\mathbf{100}\mathbf{‐}\mathbf{A}}$current is passed across the tube that is perpendicular to alocalid="1654594587153" ${}^{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{T}}$magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.)

Short answer

The force exerted on the water tube is${}^{\mathbf{50}\mathbf{}\mathbf{N}}$

Step-by-step solution

Given information

The current of the lightning bolt${}^{\mathbf{I}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{}\mathbf{A}}$

The earth’s magnetic field is${}^{\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{T}}$

The diameter of the tube is${}^{\mathbf{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{CM}\mathbf{}\mathbf{\left(}\frac{1m}{100\mathrm{cm}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}}\mathbf{}$

Define magnetic force on

The magnetic force on a current-carrying conductor can be determined using the right-hand rule-1 such that, if fingers point towards the magnetic field and the thumb is in the direction of current then the palm will not be in the direction of magnetic force.

Quantitatively magnetic force can be estimated using the equation,

$\mathbf{F}\mathbf{=}\mathbf{ILB}\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$..........................................( 1 )

Evaluating the force exerted

The direction of the current and the magnetic field is said to be perpendicular, therefore substituting the given data in equation (1) is,

$$\begin{gathered} \mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{m}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{100}\mathbf{}\mathbf{A}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{T}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{sin}\mathbf{\left(}{\mathbf{90}}^{\mathbf{°}}\mathbf{\right)} \\ \mathbf{=}\mathbf{}\mathbf{50}\mathbf{}\mathit{N} \end{gathered}$$

Therefore, the force exerted on the water tube is${}^{\mathbf{50}\mathbf{}\mathbf{N}}$