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प्रश्न
Two long parallel wires, both carrying current I directed into the plane of the page, are separated by a distance R. Show that at a point P equidistant from the wires and subtending an angle θ from the plane containing the wires, the magnitude of the magnetic field is B = `(mu_0)/pi "I"/"R"` sin 2θ. What is the direction of the magnetic field?
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उत्तर
`vec"B"_1 and vec"B"_2` are the magnetic fields in the plane of the page caused by currents in wires 1 and 2, respectively, in the diagram above.
The right-hand grip rule gives them directions: `vec"B"_1` is perpendicular to AP and forms an angle Φ with the horizontal. `vec"B"_2` is perpendicular to BP and also forms an angle Φ with the horizontal.
AP = BP = a = `("R"//2)/(cos theta)`
and `"B"_1 = "B"_2 = mu_0/(4pi) (2"I")/"a" = mu_0/(4pi) (2"I"(2 cos theta))/"R"`
`= mu_0/pi "I"/"R" cos theta`
Because the vertical components cancel each other out, the magnitude of the net magnetic induction at P is
`"B"_"net" = 2"B"_1 cos phi = 2 "B"_1 cos (90^circ - theta) = 2"B"_1 sin theta`
`= 2 (mu_0/pi "I"/"R" cos theta) sin theta = mu_0/pi "I"/"R" sin 2theta` as required.
`vec"B"_"net"` is shown in the figure in the plane parallel to that of the wires and to the right.
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