Chapter 7: Q7.34P (page 336)

Question: A fat wire, radius a, carries a constant current I , uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

Short Answer

Expert verified

Answer

The magnetic field in the gap is $B = \frac{μ_{0} I s}{2 π a^{2}}$ .

Step by step solution

Write the given data from the question.

The radius of the wire is a .

The constant current is the wire is I.

The gap between the wire is w<<a.

Step 2: Determine the formulas to calculate the magnetic field in the gap.

The expression for the current density is given as follows.

$J = \frac{I}{A}$

Here, A is the area of the wire.

Calculate the magnetic field in the gap.

Consider the figure of the wire with the gap.

Calculate the current density.

Substitute $π a^{2}$ for A into equation (1).

$J_{d} = \frac{I}{π a^{2}} \hat{z}$

The expression for the magnetic field from the Ampere’s law is given by,

$\oint B \cdot d l = μ_{0} I_{d_{e n c}} B (2 π s) = μ_{0} I_{d_{e n c}} B = \frac{μ_{0} I_{d_{e n c}}}{2 π s}$

Substitute $J_{d} (π s^{2})$ for $I_{d_{e n c}}$ into above equation.

$B = \frac{μ_{0} J_{d} (π s^{2})}{2 π s} B = \frac{μ_{0} J_{d} s}{2}$

Substitute $\frac{I}{π a^{2}}$ for $J_{d}$ into above equation.

$B = \frac{μ_{0} \frac{I}{π a^{2}} s}{2} B = \frac{μ_{0} I s}{2 π a^{2}}$

Hence the magnetic field in the gap is $B = \frac{μ_{0} I s}{2 π a^{2}}$ .

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Question: A fat wire, radius a, carries a constant current I , uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

Short Answer

Step by step solution

Write the given data from the question.

Step 2: Determine the formulas to calculate the magnetic field in the gap.

Calculate the magnetic field in the gap.

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