. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?


  1. Answer:

    magnitude of net magnetic field at given point is

    B = 5 \times 10^{-6} T


    As we know that magnetic field due to a long current carrying wire is given as

    B = \frac{\mu_o i}{2\pi r}

    here we we will find the magnetic field due to wire which is along x axis is given as

    i = 30 A

    r = 2 m

    now we have

    B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}

    B_1 = 3\times 10^{-6} T into the plane

    Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

    i = 40 A

    r = 2 m

    now we have

    B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}

    B_2 = 4\times 10^{-6} T along + x direction

    Since the two magnetic field is perpendicular to each other

    So here net magnetic field is given as

    B = \sqrt{B_1^2 + B_2^2}

    B = 5 \times 10^{-6} T


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