Answer:
The magnetic field is [tex]B = 3.87 *10^{-6} \ T[/tex]
Direction is inward
Step-by-step explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The radius of BC is [tex]r_{BC} = 30 \ cm = 0.30 \ m[/tex]
The radius of DA is [tex]r _{DA} = 20 \ cm = 0.20 \ m[/tex]
The length of CD is [tex]r_{CD} = 10 cm = 0.1 \ m[/tex]
The length of AB is [tex]r_{AB} = 10 cm = 0.1 \ m[/tex]
The current is [tex]I = 11.4A[/tex]
The magnetic field is mathematically represented as
[tex]B = B_{BC} + B_{DA} + B_{CD} + B_{AB}[/tex]
[tex]B = \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}][/tex]
Where
[tex]\theta_{BC} = \theta_{DA} = \frac{2\pi}{3}[/tex]
Where [tex]\frac{2 \pi}{3} = 120^o[/tex]
[tex]\theta_{CD} = \theta_{AB} = 0^o[/tex]
so
[tex]B = \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}][/tex]
[tex]B = 3.87 *10^{-6} \ T[/tex]
The direction is into the page
This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward
This is because according to Fleming's Left Hand Rule the direction of current is perpendicular to the direction of magnetic field so since current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction
The magnitude of the magnetic field produced by a current at point P.
[tex]11.5\ A[/tex] in the wire is [tex]4.03 \times 10^{-6}\ T[/tex].
here, current
[tex]\to I = 11.5\ A[/tex]
[tex]\to r_1 = 0.3 \ m\\\\\to r_2 = 0.2\ m\\\\[/tex]
Due to the arc of circles, the magnetic field generated from straight wires is zero.
[tex]\to B= (\frac{\theta}{360}) \times \mu_0\times \frac{I}{2} \times R\\\\[/tex]
the magnetic field's net effect
[tex]\to B = (\frac{\theta}{360}) \times \mu_0 \times \frac{I}{2} \times ( \frac{1}{r_2} -\frac{1}{r_1})\\\\\to B = ( \frac{120}{360}) \times ( 1.26 \times 10^{-6} \times \frac{11.5}{2}) \times ( \frac{1}{0.2} - \frac{1}{0.3})\\\\\to B = 4.03 \times 10^{-6}\ \ T\\\\[/tex]
The magnitude of the magnetic field produced by a current at point P.
[tex]11.5\ A[/tex] in the wire is [tex]4.03 \times 10^{-6}\ T[/tex]
Find out more about the magnetic field here:
brainly.com/question/14848188
