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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=渭0Is2蟺a2.

Step by step solution

01

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.

02

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

The expression for the current density is given as follows.

J=IA

Here, A is the area of the wire.

03

Calculate the magnetic field in the gap.  

Consider the figure of the wire with the gap.

Calculate the current density.

Substitute蟺a2 for A into equation (1).

Jd=I蟺a2z^

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

鈭�B路dl=渭0IdencB2蟺s=渭0IdencB=渭0Idenc2蟺s

Substitute Jd蟺s2 for Idencinto above equation.

B=渭0Jd蟺s22蟺sB=渭0Jds2

SubstituteI蟺a2forJd into above equation.

B=渭0I蟺a2s2B=渭0Is2蟺a2

Hence the magnetic field in the gap isB=渭0Is2蟺a2.

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