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divide the current in two parts say I1 and I2 in the loop
applying BIOT savarts law,
dB1=u°/4π× I1dl/r^2
= u°/4π× I1l1/r^2
dB2= u°/4π×I2dl/r^2
= u°/4π×I2l2/r^2
now
V=IR
V=I1R1=I2R2(since they are in parallel)
I1×l1=I2×l2
(rho and area cancel out)
so, Bnet=dB1-dB2
=u°/4π×I1l1/r^2- u°/4π×I2dl/r^2
=0
Hence proved
divide the current in two parts say I1 and I2 in the loop
applying BIOT savarts law,
dB1=u°/4π× I1dl/r^2
= u°/4π× I1l1/r^2
dB2= u°/4π×I2dl/r^2
= u°/4π×I2l2/r^2
now
V=IR
V=I1R1=I2R2(since they are in parallel)
I1×l1=I2×l2
(rho and area cancel out)
so, Bnet=dB1-dB2
=u°/4π×I1l1/r^2- u°/4π×I2dl/r^2
=0
Hence proved
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