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Suppose the current density changes slowly enough that we can (to good approximation) ignore all higher derivatives in the Taylor expansion

J(tr)=J(t)+(tr-t)J(t)+…

(for clarity, I suppress the r-dependence, which is not at issue). Show that a fortuitous cancellation in Eq. 10.38 yields

B(r,t)=μ04π∫J(r',t)×r^r2db'.

That is: the Biot-Savart law holds, with J evaluated at the non-retarded time. This means that the quasistatic approximation is actually much better than we had any right to expect: the two errors involved (neglecting retardation and dropping the second term in Eq. 10.38 ) cancel one another, to first order.

Short Answer

Expert verified

The fortuitous cancellation in Eq. 10.38 yields Br,t=μ04π∫Jr',t×r^r2db'.

Step by step solution

01

Expression for the current density and magnetic field strength.

Write the expression for the current density.

J(tr)=J(t)+(tr-t)+J(t)+…

Here, J is the current density andtr is the retarded time.

Write the expression for the magnetic field strength.

B(r,t)=μ04π∫[Jr',trr2+Jr,trcr]×r^db' …… (1)

Here, B is the magnetic field,μ0 is the permeability of free space and c is the speed of light.

02

Prove B(r,t)=μ04π∫[Jr',t×r^r2]db' :

Write the expression for the retarded time.

tr=t-rc

Substitute the value oftr in equation (1).

Br,t=μ04π∫Jr',tr2+tr-tJtr2+Jr',trcr×r^db'Br,t=μ04π∫1r2Jr',t+tr-tJt+rcJr',tr×r^db'Br,t=μ04π∫1r2Jr',t-rcJr',t+rcJr',tr1×r^db'Br,t=μ04π∫Jr',t×r^r2db'

Therefore, the fortuitous cancellation in Eq. 10.38 yields

Br,t=μ04π∫Jr',t×r^r2db'.

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