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Chapter 15: Problem 86

Two long wires carrying current are kept crossed (not joined at \(O\) ). The locus where magnetic field is zero is (A) \(I_{1}=\frac{x}{y} I_{2}\) (B) \(I_{1}=\frac{y}{x} I_{2}\) (C) \(x=y\) (D) \(x=-y\)

Short Answer

Expert verified
The short answer is: \(I_1 = \frac{x}{y} I_2\), which corresponds to option (A).

Step by step solution

01

Determine the magnetic field due to each wire individually.

To find the locus of points where the magnetic field is zero, we first need to calculate the magnetic field generated due to each wire at a point in that vicinity. Using the Biot-Savart law, the magnetic field generated due to a wire carrying a current \(I\) at a distance \(r\) is given by: \[B = \frac{\mu_0 I}{2 \pi r}\] where \(\mu_0\) is the permeability of free space.
02

Identify the points where the magnetic fields will cancel out.

At the locus of points where the magnetic field is zero, the magnetic fields due to wires carrying \(I_1\) and \(I_2\) should be equal and opposite. Let's denote the distance between point O and the point where the magnetic field becomes zero as \(x\) and \(y\) along the \(I_1\) and \(I_2\) wire directions, respectively. Using the formula for the magnetic field, we have: \[\frac{\mu_{0}I_{1}}{2\pi x} = \frac{\mu_{0}I_{2}}{2\pi y}\]
03

Solve for the locus equation.

To find the relation between \(x\), \(y\), \(I_1\), and \(I_2\), we can divide the two equations: \[\frac{\frac{\mu_{0}I_{1}}{2\pi x}}{\frac{\mu_{0}I_{2}}{2\pi y}} = 1\] The \(\mu_0\) and \(2\pi\) terms cancel out on both sides, and we are left with: \[\frac{I_{1}y}{I_{2}x} = 1\] Rearranging terms, we get the locus equation: \[\boxed{I_1 = \frac{x}{y} I_2}\] Hence, the correct answer is (A).

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Most popular questions from this chapter

A particle of mass \(M\) and charge \(Q\) moving with velocity \(\vec{v}\) describe a circular path of radius \(R\) when subjected to a uniform transverse magnetic field of induction \(B\). The work done by the field when the particle completes one full circle is \([\mathbf{2 0 0 3}]\) (A) \(\left(\frac{M v^{2}}{R}\right) 2 \pi R\) (B) Zero (C) \(B Q 2 \pi R\) (D) \(B Q v 2 \pi R\)

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