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Q10PE

Expert-verifiedFound in: Page 860

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Integrated Concepts Referring to the situation in the previous problem: (a) What current is induced in the ring if its resistance is 0.0100 Ω? (b) What average power is dissipated? (c) What magnetic field is induced at the center of the ring? (d) What is the direction of the induced magnetic field relative to the MRI’s field?**

- The value of current (I) is \({\rm{0}}{\rm{.304\;A}}\).
- The potential difference in current will be dissipated power is \(9.24 \times {\rm{1}}{{\rm{0}}^{ - 4}}\;{\rm{W}}\).
- The magnetic field induced at the center of rings is \({\rm{17}}{\rm{.4}}\;\mu {\rm{T}}\).
- The direction of the induced magnetic field is opposite

**The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. In a magnetic field, a moving charge experiences a force that is perpendicular to both its velocity and the magnetic field.**

**(a)**

The resistance is given, and we've just determined the electromotive force or potential difference \({\rm{\varepsilon }} = {\rm{\Delta V}}\).

In this case, we can simply determine the current using Ohm's law as follows:

\(\begin{aligned}{c}{\rm{I}} &= \frac{{\Delta {\rm{V}}}}{{\rm{R}}}\\ &= \frac{{3.04 \times {{10}^{ - 3}}\;{\rm{V}}}}{{{\rm{0}}{\rm{.0100}}\;\Omega }}\\ &= {\rm{0}}{\rm{.304\;A}}\end{aligned}\)

Therefore, the value of the current is \({\rm{0}}{\rm{.304\;A}}\).

**(b)**

The product of the potential difference with the current will be the dissipated power. This results in

\(\begin{aligned}{c}{\rm{P}} &= {\rm{VI}}\\ &= \left( {3.04 \times {\rm{1}}{{\rm{0}}^{ - {\rm{3}}}}\;{\rm{V}}} \right) \times {\rm{0}}{\rm{.304}}\;{\rm{A}}\\ &= 9.24 \times {\rm{1}}{{\rm{0}}^{ - 4}}\;{\rm{W}}\end{aligned}\)

Therefore, the potential difference in current will be dissipated power is \({\rm{0}}{\rm{.9\;mW}}\).

**(c)**

The magnetic field created at the ring's center is known to have a magnitude of

\({\rm{B = }}\frac{{{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{2R}}}}\)

As a result of which we may calculate it as

\(\begin{aligned}{c}{\rm{B}} &= \frac{{\left( {{\rm{4\pi }} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}\;{{T \cdot m} \mathord{\left/

{\vphantom {{T \cdot m} A}} \right.

\kern-\nulldelimiterspace} A}} \right)\; \times \left( {{\rm{0}}{\rm{.304}}\;{\rm{A}}} \right)}}{{{\rm{2}} \times \left( {{\rm{0}}{\rm{.011}}\;{\rm{m}}} \right)}}\\ &= 17.4 \times {\rm{1}}{{\rm{0}}^{ - 6}}\;T\left( {\frac{{1\;\mu T}}{{{\rm{1}}{{\rm{0}}^{ - 6}}\;T}}} \right)\\ &= {\rm{17}}{\rm{.4}}\;\mu {\rm{T}}\end{aligned}\)

Therefore, the magnitude of rings at the center is \({\rm{17}}\;{\rm{\mu T}}\).

**(d)**

The direction of the induced field will be opposite to the field whose flux increase created it, according to Lenz's law.

Therefore, the direction of the induced magnetic field is the opposite.

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