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Problem 1t. A single flat loop of wire 80 em in diameter having a resistance of 800mtl is attached to 12V battery (a) 6 pts)
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Answer #1

here ,

the diameter of loop , d = 10 cm

radius , r = d/2 = 5 cm = 0.05 m

resistance , R = 800 mohm = 0.8 ohm

potential difference , V = 12 V

a)

the current in the wire , I = V/R

I = 12 /0.8 A = 15 A

the direction of current is CLOCKWISE

b)

the magnitude of field at the center , B = u0 * I /( 2 * r)

B = 1.26 * 10^-6 * 15 /( 2 * 0.05)

B = 1.89 * 10^-4 T

using right hand thumb rule

the direction of field is Into the page

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