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Chapter 15: Problem 74

Three long straight wires are connected parallel to each other across a battery of negligible internal resistance. The ratio of their resistances are \(3: 4: 5\). What is the ratio of distances of middle wire from the others if the net force experienced by it is zero? (A) \(4: 3\) (B) \(3: 1\) (C) \(5: 3\) (D) \(2: 3\)

Short Answer

Expert verified
The ratio of distances of the middle wire from the others for the net force experienced by it to be zero is \(5:3\), which corresponds to answer choice (C).

Step by step solution

01

Identify Known Values

Given the resistance ratios, the ratio of currents in the three wires is the inverse of the ratio of resistances. Accordingly, the ratio of currents in the three wires is given by, \(I_1:I_2:I_3 = 1/3 : 1/4 : 1/5\)
02

Express the Magnetic Force

The magnetic force between the wires can be expressed using Ampere's Law as: \(F = \dfrac{kI_1I_2}{d}\), where \(k\) is the proportionality constant, and \(d\) is the distance between the wires. Assuming the middle wire has current \(I_2\), the magnetic forces between the first and second wires and between the second and third wires are: \(F_{12} = \dfrac{kI_1I_2}{d_1}\) and \(F_{23} = \dfrac{kI_2I_3}{d_2}\) For the net force acting on the middle wire to be zero, \(F_{12} = F_{23}\)
03

Equate the Forces

Equate the magnetic forces acting on the middle wire, \(\dfrac{kI_1I_2}{d_1} = \dfrac{kI_2I_3}{d_2}\)
04

Solve for the Distance Ratio

Divide both sides of the equation by \(k I_2\) and substitute the given current ratios: \(\dfrac{I_1}{d_1} = \dfrac{I_3}{d_2} \) \(\dfrac{\frac{1}{3}}{d_1} = \dfrac{\frac{1}{5}}{d_2}\) Now, multiply both sides of the equation by 15 to eliminate fractions: \(5d_2 = 3d_1\) Dividing both sides by 3, we get \(d_2 = \dfrac{3}{5}d_1\), which gives us the ratio of distances \(d_1 : d_2 = 5:3\) So, the ratio of distances of the middle wire from the others is \(5:3\), matching answer choice (C).

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Most popular questions from this chapter

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]

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