A straight segment OP of length L of a circuit carrying current I ampere is placed along x-axis. Two infinitely long straight wires A and B each extending from z=−∞ to +∞ are fixed at y=−l meter to y=+l meter respectively. Wires A and B, each carry current I ampere along positive z-axis. Given that O is origin of the coordinate system. The magnitude of force on segment OP is

Net B at distance x from O along x-axisμ02πrIr2cosθ⇒F=∫dF=∫Iμ02πIr2cosθdx=μ02πI2ln(1+L2l2)

A straight segment OP of length L of a circuit carrying current I ampere is placed along x-axis. Two infinitely long straight wires A and B each extending from z=−∞ to +∞ are fixed at y=−l meter to y=+l meter respectively. Wires A and B, each carry current I ampere along positive z-axis. Given that O is origin of the coordinate system. The magnitude of force on segment OP is

Net B at distance x from O along x-axisμ02πrIr2cosθ⇒F=∫dF=∫Iμ02πIr2cosθdx=μ02πI2ln(1+L2l2)

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A straight segment OP of length L of a circuit carrying current I ampere is placed along x-axis. Two infinitely long straight wires A and B each extending from $z = - \infty$ to $+ \infty$ are fixed at $y = - l$ meter to $y = + l$ meter respectively. Wires A and B, each carry current I ampere along positive z-axis. Given that O is origin of the coordinate system. The magnitude of force on segment OP is

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$\frac{μ_{0}}{π} I^{2} l n (1 + \frac{L^{2}}{l^{2}})$

$\frac{μ_{0}}{2 π} I^{2} l n (1 + \frac{L^{2}}{l^{2}})$

$\frac{A_{0}}{π} I^{2} l n (1 + \frac{2}{L^{2}})$

$\frac{μ_{0}}{2 π} I^{2} l n (1 + \frac{l^{2}}{L^{2}})$

answer is B.

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Detailed Solution

Net B at distance x from O along x-axis

$\frac{μ_{0}}{2 π r} \frac{I}{r} 2 \cos θ \Rightarrow F = \int d F = \int I \frac{μ_{0}}{2 π} \frac{I}{r} 2 \cos θ d x = \frac{μ_{0}}{2 π} I^{2} l n (1 + \frac{L^{2}}{l^{2}})$

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