Two square conducting loops carry currents of 5.4 A (on the left) and 3.4 A (on

Question

Two square conducting loops carry currents of 5.4 A (on the left) and 3.4 A (on the right) as shown in Fig. 30-49 (please use these values of the currents instead of those shown in the figure). What is the value of the integral of B?l (sign and magnitude) for each of the two closed paths shown?

Path 1: (  )*10^-6 T m

Path 2: (  )*10^-6 T m

Two square conducting loops carry currents of 5.4 A (on the left) and 3.4 A (on the right) as shown in Fig. 30-49 (please use these values of the currents instead of those shown in the figure). What is the value of the integral of B?l (sign and magnitude) for each of the two closed paths shown? Path 1: ( )*10^-6 T m Path 2: ( )*10^-6 T m

Explanation / Answer

this follows from amperes law, which states that for any closed surface, line integral of B.dl = uo times the currrent enclosed within that surface

so B.dl = u0i

for closed surface in path1, net current i = 5-3 = 2A (since both are opposite to each other

thus B.dl = 4pi*10^-7 * 2 = 2.512*10^-6 Tm

for closed surface in path 2, net current i = 5-5+3= 3A

thus B.dl = 4pi*10^-7*3 =3.768*10^-6 Tm

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