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Chapter 13: Problem 1930

Two thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$

Short Answer

Expert verified
The correct answer is (b) \(\left[\left(\mu_{0} I^{2}\right) /(2 \pi \mathrm{y})\right]\).

Step by step solution

01

Write down the formula for the force between two current-carrying wires

The formula for the force F per unit length (F/L) between two current-carrying wires is given by: \(\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}\) where - \(\mu_0\) is the magnetic constant (permeability of free space), - \(I_1\) and \(I_2\) are the currents in the two wires, - \(d\) is the distance between the wires. Since both wires have the same current I, we replace \(I_1\) and \(I_2\) with I in the formula: \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi d}\)
02

Find the correct substitution for the distance

Given that the distance between the wires is y, we need to substitute the variable d with y to get the formula: \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi y}\)
03

Compare the result with the given options

The formula we found is \(\frac{F}{L}=\frac{\mu_0 I^2}{2\pi y}\), which matches the option (b). Therefore, the correct answer is (b) \(\left[\left(\mu_{0} I^{2}\right) /(2 \pi \mathrm{y})\right]\).

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Most popular questions from this chapter

A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)

An electron moving with a speed \(y_{0}\) along the positive \(x\) -axis at \(\mathrm{y}=0\) enters a region of uniform magnetic field \(\mathrm{B}^{-}=-\mathrm{B}_{0} \mathrm{k} \wedge\) which exists to the right of y-axis. The electron exits from the region after some time with the speed at co-ordinate y then.

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 proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)
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