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Chapter 16: Problem 7

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(I_{1}\) and \(I_{2}\), respectively. 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 \(A B C D\) 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) \(\frac{\mu_{0}}{2 \pi a}\left(\frac{I_{1} I_{2}}{I_{1}+I_{2}}\right)\)

Short Answer

Expert verified
The short answer is: \(B = \frac{\mu_{0}}{2 \pi a}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}\), which corresponds to option (C).

Step by step solution

01

Find the magnetic field at point P due to the conductor AOB

We will use the formula for the magnetic field due to an infinite straight conductor carrying a current I: \[B = \frac{\mu_0 I}{2 \pi r}\] where \(B\) is the magnetic field strength, \(\mu_0\) is the permeability of free space (\(4 \pi x 10^{-7} Tm/A\)), \(I\) is the current flowing through the conductor, and \(r\) is the perpendicular distance from the conductor to the point where magnetic field strength is to be calculated. Since we need to find the magnetic field at point \(P\) due to conductor \(AOB\), we have \(I = I_1\) and \(r = a\) in our formula: \[B_{1} = \frac{\mu_0 I_1}{2 \pi a}\]
02

Find the magnetic field at point P due to the conductor COD

Similarly, we will use the same formula to find the magnetic field at point \(P\) due to conductor \(COD\). This time, \(I = I_2\) and \(r = a\) in our formula: \[B_{2} = \frac{\mu_0 I_2}{2 \pi a}\]
03

Find the resultant magnetic field at point P

Since conductors \(AOB\) and \(COD\) are perpendicular to each other, the magnetic fields due to these conductors at point \(P\) will be perpendicular as well. Therefore, we can use vector addition to find the resultant magnetic field: \[B = \sqrt{B_1^2 + B_2^2}\] Using the expressions for \(B_1\) and \(B_2\) that we found in steps 1 and 2, we get: \[B = \sqrt{\left(\frac{\mu_0 I_1}{2 \pi a}\right)^2 + \left(\frac{\mu_0 I_2}{2 \pi a}\right)^2}\]
04

Simplify the expression for the resultant magnetic field

First, we'll factor out the common term \(\frac{\mu_0}{2 \pi a}\): \[B = \frac{\mu_0}{2 \pi a} \sqrt{I_1^2 + I_2^2} \] Comparing this expression to the given choices in the problem, we find that the correct answer is: \(B = \frac{\mu_{0}}{2 \pi a}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}\) Hence, the correct answer is option (C).

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Most popular questions from this chapter

In a uniform magnetic field of induction \(B\), a wire in the form of a semicircle of radius \(r\) rotates about the diameter of the circle with an angular frequency \(\omega\). The axis of rotation is perpendicular to the field. If the total resistance of the circuit is \(R\), the mean power generated per period of rotation is (A) \(\frac{(B \pi r \omega)^{2}}{2 R}\) (B) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\) (C) \(\frac{B \pi r^{2} \omega}{2 R}\) (D) \(\frac{\left(B \pi r \omega^{2}\right)^{2}}{8 R}\)

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