The figure below shows a head-on view of two parallel currents. i_1runs out of t

Question

The figure below shows a head-on view of two parallel currents. i_1runs out of the board and i_2 runs into the board. Assume that the currents are very long and straight and that the magnitude of i_2 is twice that of i_1. Determine the net magnetic field at point #5 (the point marked halfway between the two currents). B^vector = [-mu_0 i_1/2 pi d + mu_0 i_2/2 pi d]y cap B^vector = [mu_0 i_1/2 pi d + mu_0 i_1/pi d]y cap B^vector = [mu_0 i_1/2 pi d + mu_0 i_1/pi d]y cap B^vector = -[mu_0 i_1/2 pi d + mu_0 i_2/2pi d]y cap None of the above.

Explanation / Answer

magnetic field due to a long current carrying wire is given by
B = 2ki/d [ where i is the current running through the wire and d is the distance from the wire]
from the right hand rule we can see that the magnetic fields due to both the wires at the poiint of interest will point in the +ve y diorection
so, magnetic field due to left wire = B1 = 2ki1/d
so, magnetic field due to right wire = B2 = 2ki2/d
netmagnetic field B = B1 + B2 = 2ki1/d + 2ki2/d [ here k = mu/4pi]
B = mu*i1/2*pid + mu*i2/2*pi*d
   but i2 = 2i1
   so. B = mu*i1/2*pi*d + mu*i1/pi*d

   so the answer is second option

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