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Graph of force per unit length between two long parallel current carrying conductors and the distance between them (a) Straight line (b) Parabola (c) Ellipse (d) Rectangular hyperbola

Short Answer

Expert verified
The force per unit length between two long parallel current-carrying conductors is given by f = I₂ × \(\frac{\mu_0 I_1}{2 \pi r}\), where f is the force per unit length, I₁ and I₂ are the currents in the conductors, and r is the distance between the conductors. Since f is inversely proportional to r, the graph shape is a rectangular hyperbola. Thus, the correct answer is (d) Rectangular hyperbola.

Step by step solution

01

Determine the magnetic field created by one conductor

To calculate the magnetic field (B) created by one conductor at a distance (r) from it, we can use Ampere's law. Let's consider a current I₁ flowing through the first conductor. The formula for the magnetic field due to this current is given by: B = \(\frac{\mu_0 I_1}{2 \pi r}\) where \(\mu_0\) is the permeability of free space. Now let's find the force acting on the second conductor due to this magnetic field.
02

Calculate the force on the second conductor

Now, we consider a current I₂ flowing through the second conductor. The force (F) on this conductor can be calculated using the formula: F = I₂ × (L × B) where L is the unit length of the conductor, which can be taken as 1 meter. Plugging in the formula for B from Step 1 into this equation, we get: F = I₂ × \(\frac{\mu_0 I_1}{2 \pi r}\) Now we can find the force per unit length.
03

Calculate the force per unit length

The force per unit length (f) can be found by dividing the force F by the unit length L (which is 1 meter in our case): f = \(\frac{F}{L}\) Since L = 1, we can state that: f = I₂ × \(\frac{\mu_0 I_1}{2 \pi r}\) Now let's analyze the graph of this force per unit length formula with respect to the distance between the conductors (r).
04

Analyze the graph of force per unit length between the conductors

The formula for force per unit length is given by: f = I₂ × \(\frac{\mu_0 I_1}{2 \pi r}\) We can observe that the force per unit length (f) is inversely proportional to the distance (r) between the conductors: f ∝ \(\frac{1}{r}\) This inverse relationship indicates that the graph shape should be a rectangular hyperbola. Therefore, the correct answer is (d) Rectangular hyperbola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampere's Law
Ampere's Law is a fundamental principle in electromagnetism that explains how electric currents create magnetic fields. According to this law, the magnetic field around a conductor is directly proportional to the current passing through it. Ampere's Law can be mathematically expressed using the formula:\[ B = \frac{\mu_0 I}{2 \pi r} \]where:
  • \(B\) is the magnetic field.
  • \(\mu_0\) is the permeability of free space, a constant.
  • \(I\) is the current flowing through the conductor.
  • \(r\) is the distance from the conductor.
This formula helps us understand how the magnetic field diminishes as you move away from the conductor. It also shows that the magnetic field is stronger when the current is larger.

Ampere's Law is quite useful, especially when calculating forces between current-carrying conductors as it helps determine the magnetic influence one conductor exerts on another.
Magnetic Field due to a Current
When a current flows through a conductor, like a wire, it creates a magnetic field around it. This is an essential concept in understanding how electric currents can create forces. The direction of the magnetic field lines is given by the right-hand rule: if you wrap your right hand around the conductor with your thumb pointing in the direction of the current, your fingers will curl in the direction of the magnetic field.

The strength of this magnetic field varies with the distance from the conductor and can be calculated using the formula from Ampere’s Law:\[ B = \frac{\mu_0 I}{2 \pi r} \]Here, the field decreases as you move further away, which is why proximity to the current source is important.

Understanding this magnetic field is critical for calculating forces, especially when dealing with situations involving multiple conductors, where each one will affect the others based on the strength and direction of their respective magnetic fields.
Rectangular Hyperbola Graph
In physics, particularly when analyzing interactions between current-carrying conductors, recognizing the correct graph shape is pivotal. The force per unit length ( \(f\) ) between two parallel conductors is described by the equation:\[ f = I_2 \times \left( \frac{\mu_0 I_1}{2 \pi r} \right) \]This formula shows that \(f\) is inversely proportional to the distance ( \(r\) ) between them, meaning as one increases, the other decreases.

A rectangular hyperbola graph illustrates this relationship well, reflecting how the force diminishes with increased distance. In graphical terms, this means the graph will approach both axes but never touch them.

Such visual understanding helps in predicting how forces change with varying distances, important in practical applications like designing electrical circuits and laying out conductors to minimize interference.

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Most popular questions from this chapter

The magnetic susceptibility is negative for (a) Paramagnetic materials (b) Diamagnetic materials (c) Ferromagnetic materials (d) Paramagnetic and ferromagnetic materials

The mag. field due to a current carrying circular Loop of radius \(3 \mathrm{~cm}\) at a point on the axis at a distance of \(4 \mathrm{~cm}\) from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be (a) Only inside the pipe (b) Only outside the pipe (c) Neither inside nor outside the pipe (d) Both inside and outside the pipe

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

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