Answer:
B = 15μT
Explanation:
In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:
[tex]B=\frac{\mu_oI}{2\pi r}[/tex] (1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current
r: distance from the wire to the point in which B is calculated
In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:
[tex]B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)[/tex] (2)
I1: current of the central wire = 2.00A
I2: current of the cylindrical conductor = 3.50A
r: distance = 2.00 cm = 0.02 m
You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.
[tex]|B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T[/tex]
The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT
