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Chapter 15: Problem 30

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(i_{1}\) and \(i_{2}\). The magnitude of the magnetic induction at a point \(P\) at a distance \(a\) from the point \(O\) in a direction perpendicular to the plane \(A C B D\) is (A) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}+i_{2}\right)\) (B) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}-i_{2}\right)\) (C) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}^{2}+i_{2}^{2}\right)^{1 / 2}\) (D) \(\frac{\mu_{0}}{2 \pi a} \frac{i_{1} i_{2}}{\left(i_{1}+i_{2}\right)}\)

Short Answer

Expert verified
The short answer is: The magnitude of the magnetic induction at point \(P\) is given by \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}^{2}+i_{2}^{2}\right)^{1 / 2}\).

Step by step solution

01

Determine the magnetic induction due to conductor AOB

First, we need to find the magnetic induction at point \(P\) due to the current \(i_1\) in conductor \(AOB\). We can use the Biot-Savart law for the magnetic field due to a straight long conductor, which states that the magnetic field \(B\) at a distance \(r\) from the conductor carrying a current \(I\) is given by: \[B = \frac{\mu_0 I}{2 \pi r} \] In this case, the distance between \(AOB\) and point \(P\) is \(a\), and the current in conductor \(AOB\) is \(i_1\). So the magnetic induction at point \(P\) due to conductor \(AOB\) is: \[B_{1} = \frac{\mu_0 i_1}{2 \pi a} \]
02

Determine the magnetic induction due to conductor COD

Similarly, we can find the magnetic induction at point \(P\) due to the current \(i_2\) in conductor \(COD\). The distance between \(COD\) and point \(P\) is also \(a\), so the magnetic induction at point \(P\) due to conductor \(COD\) is: \[B_{2} = \frac{\mu_0 i_2}{2 \pi a} \]
03

Apply the superposition principle

To find the total magnetic induction at point \(P\), we apply the superposition principle, which states that the total magnetic induction is the sum of the individual magnetic inductions due to each conductor. Since the conductors are perpendicular to each other, the magnetic inductions due to \(AOB\) and \(COD\) are also perpendicular to each other. We can use the Pythagorean theorem to find the magnitude of the total magnetic induction at point \(P\): \[B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \sqrt{\left(\frac{\mu_0 i_1}{2 \pi a}\right)^{2} + \left(\frac{\mu_0 i_2}{2 \pi a}\right)^{2}}\]
04

Simplify the expression

We can simplify the expression for the total magnetic induction at point \(P\): \[B = \sqrt{\frac{\mu_0^2 i_1^2}{(2 \pi a)^2} + \frac{\mu_0^2 i_2^2}{(2 \pi a)^2}}\] \[B = \frac{\mu_0}{2 \pi a} \sqrt{i_1^2 + i_2^2}\] The correct answer is (C) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}^{2}+i_{2}^{2}\right)^{1 / 2}\).

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Most popular questions from this chapter

Two long parallel wires carrying equal currents in opposite directions are placed at \(x=\pm a\) parallel to \(v\)-axis with \(z=0\). Then (A) Magnetic field \(B_{1}\) at 1\. \(\frac{\mu_{0} i}{3 \pi a}\) origin \(O\) (B) Magnetic field \(B_{2}\) at \(P\) 2\. \(\frac{\mu_{0} i}{4 \pi a}\) \((2 a, 0,0)\) \((a, 0,0)\) (C) Magnetic field at \(M\) 3\. \(\frac{\mu_{0} i}{\pi a}\) (D) If wire carries current in 4\. Zero the same direction, then magnetic field at origin 5\. None

A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque needed to maintain the needle in this position will be (A) \(\sqrt{3} W\) (B) \(W\) (C) \(\frac{\sqrt{3}}{2} W\) (D) \(2 W\)

A triangular loop of side \(l\) carries a current \(I .\) It is placed in a magnetic field \(B\) such that the plane of the loop is in the direction of \(B\). The torque on the loop is (A) Zero (B) \(I B l^{2}\) (C) \(\frac{\sqrt{3}}{2} I B l^{2}\) (D) \(\frac{\sqrt{3}}{4} I B l^{2}\)

Relative permittivity and permeability of a material \(\varepsilon_{r}\) and \(\mu_{r}\) respectively. Which of the following value of these quantities are allowed for a diamagnetic material? (A) \(\varepsilon_{r}=0.5, \mu_{r}=1.5\) (B) \(\varepsilon_{r}=1.5, \mu_{r}=0.5\) (C) \(\varepsilon_{r}=0.5, \mu_{r}=0.5\) (D) \(\varepsilon_{r}=1.5, \mu_{r}=1.5\)

Curie temperature is the temperature above which \([2003]\) (A) a ferromagnetic material becomes paramagnetic. (B) a paramagnetic material becomes diamagnetic. (C) a ferromagnetic material becomes diamagnetic. (D) a paramagnetic material becomes ferromagnetic.
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