A long straight wire carrying current 16 ~A is bent at 90^ such that half of the wire lies along the positive x -axis and other half lies along the positive y -axis. What is the magnitude of the magnetic field at the point, r =(-2 i +0 j ) mm ? (Assume, ₀ 4 =10⁻⁷ H m ⁻¹ )

Step-by-step solution

According to the question, a long straight wire is bent at 90^ as shown in the figure, Where, current flowing through the wire, I=16 ~A r = OP =(-2 i +0 j ) mm Magnitude of the magnetic field at point P due to current carrying wire along x -axis is zero because pont P lies in the direction of conducting wire along x -axis. Magnitude of the magnetic field due to current carrying wire along y -axis at point P is given by B= ₀ 4 I r Putting the given values in above relation, we get aligned & = ₀ 4 16 2 10⁻³ [ r=2 ~mm