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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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Most popular questions from this chapter

Two parallel long wires \(\mathrm{A}\) and B carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). \(\left(\mathrm{I}_{2}<\mathrm{I}_{1}\right)\) when \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are in the same direction the mag. field at a point mid way between the wires is \(10 \mu \mathrm{T}\). If \(\mathrm{I}_{2}\) is reversed, the field becomes \(30 \mu \mathrm{T}\). The ratio \(\left(\mathrm{I}_{1} / \mathrm{I}_{2}\right)\) is (a) 1 (b) 2 (c) 3 (d) 4

A magnet of magnetic moment \(\mathrm{M}\) and pole strength \(\mathrm{m}\) is divided in two equal parts, then magnetic moment of each part will be (a) \(\mathrm{M}\) (b) \((\mathrm{M} / 2)\) (c) \((\mathrm{M} / 4)\) (d) \(2 \mathrm{M}\)

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

For the mag. field to be maximum due to a small element of current carrying conductor at a point, the angle between the element and the line joining the element to the given point must be (a) \(0^{\circ}\) (b) \(90^{\circ}\) (c) \(180^{\circ}\) (d) \(45^{\circ}\)

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