Q7PE Verify that the units of \({\rm{... [FREE SOLUTION]

Chapter 23: Q7PE (page 860)

Verify that the units of \({\rm{\Delta \Phi /\Delta t}}\) are volts. That is, show that \({\rm{1\;T \times }}{{\rm{m}}^{\rm{2}}}{\rm{/s = 1\;V}}\).

Short Answer

Expert verified

Hence, verified that the units of \({\rm{\Delta \Phi /\Delta t}}\) are volts.

Step by step solution

Definition of Volt

Electric potential, electric potential difference, and electromotive force are all measured in volts. It is named after Alessandro Volta, an Italian physicist.

Substituting the units

Keep in mind that the electromotive force is expressed as

\(\begin{aligned}{c}{\rm{\varepsilon }} &= - \frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}\\ &= \frac{{{\rm{\Delta (BA)}}}}{{\rm{t}}}\end{aligned}\)

As a result, its units will be

\(\begin{aligned}{c}{\rm{(\varepsilon )}} &= \left( {\frac{{{\rm{\Delta (BA)}}}}{{\rm{t}}}} \right)\\ &= \frac{{{\rm{T}} \cdot {{\rm{m}}^{\rm{2}}}}}{{\rm{s}}}\end{aligned}\)

Let us recall that the magnetic field unit, Tesla, is written as

\({\rm{T}} = \frac{{{\rm{N}} \cdot {\rm{s}}}}{{{\rm{C}} \cdot {\rm{m}}}}\)

If you're not sure, recall that the force acting on a charge \(q\) moving at \(v\) in a magnetic field \(B\) is given by

\(\begin{aligned}{l}{\rm{F}} &= {\rm{qvB}}\\{\rm{B}} &= \frac{{\rm{F}}}{{{\rm{qv}}}}\end{aligned}\)

Thus, substituting the units and noting that the charge over time can be substituted for the current, gives

\({\rm{T}} &= \frac{{\rm{N}}}{{{\rm{C}} \cdot {{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}}}\)

Now that we've established the magnetic field's units, we may go on.

\(\begin{aligned}{c}{\rm{(\varepsilon )}} &= {\rm{T}}\frac{{{{\rm{m}}^{\rm{2}}}}}{{\rm{s}}}\\ &= \frac{{{\rm{N}} \cdot {\rm{s}}}}{{{\rm{C}} \cdot {\rm{m}}}} \times \frac{{{{\rm{m}}^{\rm{2}}}}}{{\rm{s}}}\\ &= \frac{{{\rm{N}} \cdot {\rm{m}}}}{{\rm{C}}}\end{aligned}\)

Unit of Nm

Finally, keep in mind that the joule is the unit of N.m, as pulling a force for a particular distance produces work. This results in

\({\rm{(\varepsilon )}} = \frac{{\rm{J}}}{{\rm{C}}}\)

One joule of energy is required to transfer one coulomb of charge across the potential of one volt, therefore it's only the volt.

Therefore, theunit of a volt is,

\(\begin{aligned}{c}{\rm{(\varepsilon )}} &= \frac{{{\rm{T}} \cdot {{\rm{m}}^{\rm{2}}}}}{{\rm{s}}}\\ &= \frac{{{\rm{N}} \cdot {\rm{m}}}}{{\rm{C}}}\\ &= {\rm{V}}\end{aligned}\)