A long, cylindrical conductor of radius R carries a current I as shown in the fi
ID: 2262510 • Letter: A
Question
A long, cylindrical conductor of radius R carries a current I as shown in the figure below. The current density J, however, is not uniform over the cross-section of the conductor but is a function of the radius according to J = br, where b is a constant.
Find B=.............
when
a) r1 < R
b) r2 > R
Explanation / Answer
Use Ampere's Law to solve this question:
Surface integral (Magnetic field*ds) = ?0 * current
Current density is J = 2br^5
Plugging in Current density the equation becomes:
Magnetic field *distance= ?0 * Integral (J*dA)
distance= 2pi*r1
Magnetic field * (2pi*r1) = ?0 * Integral from r1 to 0, (J*2pi*r dr)
further more replace J, with 2br^5
Magnetic field * (2pi*r1) = ?0 * Integral from r1 to 0, ((2br^5)*2pi*r dr)
solving the integral we get:
Magnetic field * (2pi*r1) = (?0 *4*pi*b *r1^7) /7
Magnetic field = (?0 *4*pi*b *r1^7) / (7(2pi*r1^1))
Magnetic field = 5(?0 *2*b *r1^6) / (7)
Part B is solved similarly:
Magnetic field *distance= ?0 * Integral (J*dA)
distance= 2pi*r2 (*this is now r2)
Magnetic field * (2pi*r2) = ?0 * Integral from R to 0, (J*2pi*r dr)
plug in J:
Magnetic field * (2pi*r2) = ?0 * Integral from R to 0, ((2br^5)*2pi*r dr)
solving the integral we get:
Magnetic field * (2pi*r2) = (?0 4pi*b R^7) / 7
Magnetic field = (?0 4pi*b R^7) / (7 * (2pi*r2))
Magnetic field = 5(?0 2*b R^7) / (7 * (r2))
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