Question

A square ammeter has sides of length 3.00 cm. The sides of the ammeter are capable of measuring the magnetic field they are subject to. When the ammeter is clamped around a wire carrying a direct current, as shown in the figure, the average value of the magnetic field measured in the sides is 4.00 G, What is the current in the wire? A coaxial cable consists of an insulated wire carrying current I_I=4.00A surrounded by a cylindrical conductor carrying current I_2=1.50A in the oppositive direction. calculate magnetic field inside the cylindrical conductor at r_int = 0.500 cm calculate magnetic field outside the cylindrical conductor at r_ext = 1.50 cm.

Answer 1

4. Applying Ampere´s law,

$\oint \vec{B}.d\vec{l}= \mu _{0}I$ _____________________________(1)

we choose a square path,

$B4l= \mu _{0}I,\, \, \, \, \, I= \frac{4Bl}{\mu _{0}}$ _____________________(2)

using (2), the current is

$I= \frac{4\times 4\times 10^{-4}\, T\times 3\times 10^{-2}\, m}{4\pi \times 10^{-7}\frac{Tm}{A}}=0.38 \, A$

______________________________________________

5) The magnetic field is given by

$\vec{B}= \vec{B}_{1}+\vec{B}_{2}$ ,_____________________________(1)

where,

$\oint \vec{B}_{j}.d\vec{l}= \mu _{0}I_{j},\, \, \, \, \, \, j= 1,2$ _____________________(2)

The magnetic field inside is

$\oint \vec{B}_{1}.d\vec{l}= \mu _{0}I_{1},\, \, \, \, \, \, \, \vec{B}_{1}= \frac{\mu _{0}I_{1}}{2\pi r}\hat{\theta }$ _________________(3)

$\oint \vec{B}_{2}.d\vec{l}= \mu _{0}I_{2},\, \, \, \, \, \, \, \vec{B}_{2}= \frac{\mu _{0}I_{2}}{2\pi r}\left ( -\hat{\theta } \right )$ ______________(4)

substituting (3) and (4) into (1), yields (inside)

$\vec{B}_{in}= \frac{\mu _{0}}{2\pi r_{i}} \left ( I_{1}-I_{2} \right ) \hat{\theta }= \frac{4\pi\times 10^{-7} \times 2.5}{2\pi \times 0.5\times 10^{-2}\, }\, T\, \hat{\theta }= 10^{-4}\, T\, \hat{\theta }$ ____________________(5)

_________________________

The magnetic field outside the ciindrical conductor

$\vec{B}_{out}=\left (\frac{\mu _{0}I_{1}}{2\pi r_{1}}-\frac{\mu _{0}I_{2}}{2\pi r_{2}} \right ) \hat{\theta }= -0.66\times 10^{-5}\, T\, \hat{\theta }$

___________________________________________________