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Expert-verified Found in: Page 812 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # (a) What is the force per meter on a lightning bolt at the equator that carries ${}^{\mathbf{20}\mathbf{,}\mathbf{000}\mathbf{}\mathbf{A}}$ perpendicular to the Earth’s${}^{\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{‐}\mathbf{T}}$ field? (b) What is the direction of the force if the current is straight up and the Earth’s field direction is due north, parallel to the ground?

(a)The force per meter is${}^{0.6\mathrm{N}}{\mathrm{m}}}$.

(b)The direction of the magnetic force is towards the left (west).

See the step by step solution

## Step 1: Given information

The current of the lightning bolt${}^{\mathbf{I}\mathbf{=}\mathbf{20}\mathbf{,}\mathbf{000}\mathbf{}\mathbf{A}}$

The earth’s magnetic field is${}^{\mathbf{B}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{T}}$

The angle at the equator will be${}^{\mathbf{\theta }\mathbf{=}{\mathbf{90}}^{\mathbf{°}}}$

## Step 1: Given information

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,

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

## Step 3: Evaluating the force

(a)

To evaluate the force per meter, we use the equation and substitute the given data in equation (1), which results in

$\frac{\mathbf{F}}{\mathbf{L}}\mathbf{=}\mathbf{IB}\mathbf{}\mathbf{sin}\mathbf{\left(}{\mathbf{90}}^{\mathbf{°}}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left(}\mathbf{20}\mathbf{,}\mathbf{000}\mathbf{}\mathbf{A}\mathbf{\right)}\mathbf{×}\mathbf{\left(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{×}\mathbf{1}{\mathbf{O}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{T}\mathbf{}\mathbf{\right)}\mathbf{×}\mathbf{1}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$

Hence, the force per meter is${}^{0.6\mathrm{N}}{\mathrm{m}}}$.

## Step 4: The direction of the magnetic force(b)

To obtain the direction of the force, we will be using the right-hand rule as stated above,

Now, adjust the thumb with the current's direction which is upwards, and then adjustthe fingers' direction to the magnetic field's direction which is towards the north (forward). Noticing that the palm will be facing towards the west(to the left) andtherefore the direction of the magnetic force will be towards the west or left. ### Want to see more solutions like these? 