Respuesta :
The magnetic field is just a unit vector that explains the electromagnetic influence on electric currents, current flow, and magnetic fluids.
Magnetic field:
Current density [tex]J = ar^2[/tex]
value of the constant is= [tex]28.5 \ \frac{A}{mm^4}[/tex]
radius = 5.20 mm
magnetic permeability [tex]\mu = 4\pi \times 10^7 \ \frac{N}{A^2}[/tex]
calculating the area element for the straight circular conduction rod
:
[tex]d_A=2\pi r dr[/tex]
Calculating the current, which is carried in the rod
[tex]dI = \int dA \vec{J}[/tex]
Calculating the above equation with the limit value that is 0 to r.
[tex]I^{1}=\int_{0}^{r} ar^2 \times 2 \pi \cdot r d_r[/tex]
[tex]=2\pi a\int_{0}^{r} r^3 d_r \\\\=\frac{\pi ar^4}{2}[/tex]
The calculated current value which is carried by the rod is [tex]\boxed{\frac{\pi ar^4}{2}}[/tex]
In option (a)
calculating the magnitude of the magnetic field at the point [tex]r_1= \frac{R}{2}[/tex]
[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r}\\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}r^4)\\\\\to B \times 2\pi = \mu_{0} (\frac{\pi a}{2}r^3)\\\\\to B= \frac{\mu_{0}}{4}r^3 a\\\\[/tex]
Substituting the above value:[tex]\frac{R}{2} \ for \ r[/tex]
[tex]B= \frac{\mu_{0}}{4}(\frac{R}{2})^3 a[/tex]
[tex]B= \frac{4 \pi \times 10^{-7}}{4}(\frac{5.2 \times 10^{-3}}{2})^3 \frac{28.5}{10^{-12}}[/tex]
[tex]= \frac{4 \pi \times 10^{-7}}{4} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{1} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\=\frac{1572.87624 \times 10^{-16}}{ 10^{-12}}\\\\=0.157 \ \ T[/tex]
Thus, the magnitude of the magnetic field at the point [tex]r_1 =\frac{R}{2}[/tex] is [tex]\boxed{0.157 \ T}[/tex]
In option (b)
In this, we calculate the magnitude of the magnetic field at the point [tex]r_2= 2R[/tex]
[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r} \\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}R^4)\\\\\to B \times 2\pi 2R = \mu_{0} (\frac{\pi a}{2}R^3)\\\\\to B= \frac{\mu_{0}}{8}R^3 a\\\\[/tex]
Substituting the values
[tex]B= \frac{4 \pi \times 10^{-7}}{8}(5.2 \times 10^{-3})^3(\frac{28.5}{10^{-12}})[/tex]
[tex]= \frac{4 \times 3.14 \times 10^{-7}}{8} \times (5.2)^3 \times (10^{-3})^3 \times\frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{2} \times 140.608 \times 10^{-9} \times\frac{28.5}{10^{-12}}\\\\= 6291.50495 \times 10^{-4}\\\\= 0.629 \ \ or \ \ 0.63\\[/tex]
Thus, the magnitude of the magnetic field at the point [tex]r_2 = 2R[/tex] is [tex]\boxed{0.63 \ \ T}[/tex]
Find out more about Magnetic fields here:
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