Two very long coaxial cylindrical conductors are shown in cross-section above. T
ID: 1787265 • Letter: T
Question
Two very long coaxial cylindrical conductors are shown in cross-section above. The inner cylinder has radius a = 2 cm and caries a total current of I1 = 1.2 A in the positive z-direction (pointing out of the screen). The outer cylinder has an inner radius b = 4 cm, outer radius c = 6 cm and carries a current of I2 = 2.4 A in the negative z-direction (pointing into the screen). You may assume that the current is uniformly distributed over the cross-sectional area of the conductors. What is Bx, the x-component of the magnetic field at point P which is located at a distance r = 5 cm from the origin and makes an angle of 30o with the x-axis? Bx =
Explanation / Answer
calculate total current enclosed by ardius r
Total cuurent enclosed = +1.2 A - 2.4*(pi*r^2 - pi*b^2)/(pi*c^2 - pi*b^2)
= +1.2 - 2.4*(5^2 - 4^2)/(6^2 - 4^2)
=+1.2 - 2.4 *0.45
=+1.2 - 1.08
=0.12 A
+ Sign shows current is in + ve z direction
Magnitude of B = miuo*Ienc/(2*pi*r)
Put values
= (1.2566*10^-6)*(0.12) / (2*pI*0.05)
{since 5 cm = 0.05 m}
=4.8*10^-7 T
Then direction can be found out bu right hand thumb rule
It will be perpendicular to r
Angle = 30+90=120 degree from + ve x axis
x componenet = 4.8*10^-7 * cos 120
=4.8*10^-7*(-0.5)
= -2.4*10^-7 T
Y componenet = 4.8*10^-7 * sin 120
=4.8*10^-7*(0.8660)
= 4.1568*10^-7 T
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