Chapter 1: Problem 6
A magnetic induction \(\mathrm{B}\) is generated by electric current in a ring of radius \(b\).Show that the magnitude of the vector potential \(\mathrm{A}(\mathrm{B}=\nabla \times \mathrm{A})\) at the ring $$ |\mathbf{A}|=\frac{\varphi}{2 \pi R}, $$ where \(\varphi\) is the total magnetic flux passing through the ring. Note. A is tangential to the ring.
Short Answer
Step by step solution
Understand the Given Information
Recall the Definition of Magnetic Flux
Understand Vector Potential and its Relationship with B
Express Vector Potential in Terms of Magnetic Flux
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnetic Flux
- Magnetic flux links the field strength with a specified area.
- It is measured in Weber (Wb).
- A consistent field over an area simplifies calculation: \( \varphi = B \times A \) if field \( B \) is uniform and perpendicular to \( A \).
Magnetic Field
- It is represented as \( \mathbf{B} \) and measured in teslas (T).
- The magnetic field lines emerge from the north pole and segment into the south pole of a magnet.
- In a ring of current, the field is produced by moving charges, creating circular lines of force around the wire.
Ring of Current
- Current in the ring creates a field akin to how magnets produce magnetic field lines.
- The geometric symmetry simplifies how fields and potential are calculated.
- Electric charge motion within the ring leads to magnetic induction \( \mathbf{B} \), affecting the field lines.
Biot-Savart Law
- It gives a direct relationship between current elements and produced magnetic fields.
- For an infinitesimally small current segment \( I d\mathbf{l} \), the field \( d\mathbf{B} \) at a point is \( d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{r}}{r^3} \), where \( \mathbf{r} \) is displacement from current to point.
- The law encapsulates the principle that current-carrying wires are magnetic field sources.