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Chapter 13: Problem 1944

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

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) \(75 \mu \mathrm{T}\)

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.(B) Ammeter (Q) Moderate resistance (C) Voltmeter (R) High, Low or moderate resistance (D) Avometer (S) High resistance (a) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (b) $\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$ (c) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}$ (d) $\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}$

If two streams of protons move parallel to each other in the same direction, then they (a) Do not exert any force on each other (b) Repel each other (c) Attract each other (d) Get rotated to be perpendicular to each other.

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be \(2 \mathrm{sec}\). The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be \(2 \mathrm{sec}\). Then the angle of dip is (a) \(0^{\circ}\) (b) \(30^{\circ}\) (c) \(45^{\circ}\) (d) \(90^{\circ}\)
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