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Assume that *I* = 4.6A and *d* = 2.5cm .

– What is the strength of the magnetic field at the center of the loop in the figure?

– What is the direction of the magnetic field at the center of the loop?

r = d/2 = 2.5/2 = 1.25 cm = 0.0125 cmBnet = B(stright wire) + B(loop)= mue*I/(2*pi*r) + mue*I/(2*r)= 4*pi*10^-7*4.6/(2*pi*0.0125) +4*pi*10^-7*4.6/(2*0.0125)

= 3.048*10^-4 TYou can find more information here

radius ,

r = 2.5/2

r = 1.25cm = 0.0125m

Strength of field, now

B = u0*I/ (2*pi*r), + u0*I/ (2r).

B= u0*4.6/(2*.0125) * (1/pi + 1)

B = 3.085 *10

^{-4 }TThe strength of the magnetic field at the center of the loop is 3.085 *10^{-4}Tthe direction of field is INTO THE PAGE