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Numericals · Q23

Q.Two long parallel wires, both going into the plane of the paper, are separated by a distance R and carry a current I each, in the same direction. Show that the magnitude of the magnetic field at a point P, equidistant from the wires, and subtending an angle θ\theta at P between the lines joining P to each wire, is B=μ0IπRsin⁡θB=\frac{\mu_0 I}{\pi R}\sin\theta. What is the direction of the magnetic field?

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Let the two wires, both carrying current I into the plane of the paper, be separated by R, and let P be a point equidistant from both wires, with θ\theta the angle subtended at P by the two wires (i.e. the angle between the line from P to wire 1 and the line from P to wire 2). Each wire, at its own distance ρ\rho from P, produces a field of magnitude μ0I/(2πρ)\mu_0I/(2\pi\rho) directed perpendicular to the line from that wire to P (Section 10.10.1). Setting up the geometry with P on the perpendicular bisector of the two wires (the locus of all equidistant points), and resolving each wire's field into a component along the line JOINING the two wires and a component along the perpendicular bisector: by the mirror symmetry of the configuration, the two perpendicular-bisector components are equal and OPPOSITE, and cancel exactly, while the two along-the-baseline components are equal and point the SAME way, and add. Carrying out this vector addition (using ρ=R/2sin⁡θ\rho=\dfrac{R/2}{\sin\theta}, from the triangle formed by P and the two wires with apex angle θ\theta at P) gives a net field, along the line joining the wires, of magnitude B=μ0IπRsin⁡θB=\dfrac{\mu_0 I}{\pi R}\sin\theta -- confirming the stated result. In the special case $\theta\ …

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