B due to wire (1)B1=(2πdμ0i1)i^
B due to wire (2)B2=(2πdμ0i2)(−j^)
∣Bnet∣=2πdμ0i12+i22
Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ‘ O ’ is their common point for the two. The wires carry I1 and I2 currents, respectively. Point ‘ I ’ is lying at distance ‘ d ’ from ‘ O ’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ‘ P ’ will be:
Held on 30 Apr 2014 · Verified 9 Jul 2026.
2πdμ0(I2I1)
2πdμ0(I1+I2)
2πdμ0(I12−I22)
2πdμ0(I12+I22)1/2
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