Answer:
The value is [tex]B = 0.00122 \ T [/tex]
Explanation:
The radius is [tex]r = 0.35 \ cm = 0.0035 \ m[/tex]
The current is is [tex]I = 75 \ A[/tex]
The position considered is [tex]x = 0.20 \ cm = 0.002 \ m[/tex]
Generally the magnetic field 0.20 cm from the center of the wire is mathematically represented as
[tex]B = \frac{\mu_o * I * x}{4 \pi * r^2}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
Generally
[tex]B = \frac{4\pi * 10^{-7} * 75 * 0.002 }{4* 3.142 * (0.0035)^2}[/tex]
=> [tex]B = \frac{4\pi * 10^{-7} * 75 * 0.002 }{4* 3.142 * (0.0035)^2}[/tex]
=> [tex]B = 0.00122 \ T [/tex]
The value of the megnetic field will be 0.00122 T.It is the kind of domain where the magnetic force is received.
It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.
The given data in the problem is;
r is the radius= 0.35 cm= 0.0035 m
I is the current = 75 A
x is the position considered = 0.20 cm=0.002 m
The megnetic field is found as;
[tex]\rm B= \frac{\mu_0 Ix }{4 \pi r^2} \\\\ \rm B= \frac{4 \pi \times 10^{-7} \times 75 \times 0.002 }{4 \pi (0.0035)^2} \\\\ \rm B=0.00122\ T[/tex]
Hence the value of the megnetic field will be 0.00122 T.
To learn more about the magnetic field refer to the link;
https://brainly.com/question/19542022
