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Chapter 15: Problem 33

The distance between two thin long straight parallel conducting wires is \(b\). On passing the same current \(i\) in them, the force per unit length between them will be (A) Zero (B) \(\frac{\mu_{0} i^{2}}{2 \pi b}\) (C) \(\frac{\mu_{0} i}{2 \pi b}\) (D) \(\frac{\mu_{0} i}{2 \pi}\)

Short Answer

Expert verified
The force per unit length between the two parallel conducting wires carrying the same current i is (B) \(\frac{\mu_{0} i^{2}}{2 \pi b}\).

Step by step solution

01

Find the magnetic field created by one wire

Using Ampere's law, the magnetic field, B, produced by a long straight wire carrying a current, i, at a distance, r, from the wire can be found using the formula: \(B = \frac{\mu_0 i}{2 \pi r}\), where \(\mu_0\) is the permeability of free space.
02

Determine the magnetic field at the location of the second wire

Since the wires are parallel and separated by a distance, b, we can find the magnetic field at the location of the second wire using the formula we derived in Step 1, with r equal to b: \(B = \frac{\mu_0 i}{2 \pi b}\).
03

Calculate the force per unit length between the wires

The formula for the magnetic force per unit length, F, between two long straight parallel wires carrying currents i鈧� and i鈧� and separated by a distance b can be given as: \(F = \frac{\mu_0 i_{1} i_{2}}{2 \pi b}\). In our case, both wires carry the same current i (i鈧� = i鈧� = i) so the equation becomes: \(F = \frac{\mu_0 i^2}{2 \pi b}\).
04

Choose the correct option

Comparing our result from Step 3 with the given options, we find that the force per unit length between the two parallel conducting wires is (B) \(\frac{\mu_{0} i^{2}}{2 \pi b}\).

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Most popular questions from this chapter

Current \(i\) is carried in a wire of length \(L .\) If the wire is turned into a circular coil, the maximum magnitude of torque in a given magnetic field \(B\) will be (A) \(\frac{L^{2} i B}{2}\) (B) \(\frac{L^{2} i B}{\pi}\) (C) \(\frac{L^{2} i B}{4 \pi}\) (D) \(\frac{L i^{2} B}{4 \pi}\)

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other, \(3 \mathrm{~A}\) and \(4 \mathrm{~A}\) are the currents flowing in coils, respectively. The magnetic induction in \(\mathrm{Wb} / \mathrm{m}^{2}\) at the centre of the coils will be \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}\right)\) (A) \(12 \times 10^{-5}\) (B) \(10^{-5}\) (C) \(5 \times 10^{-5}\) (D) \(7 \times 10^{-5}\)

Assertion: A conducting circular disc of radius \(R\) rotates about its own axis with angular velocity \(\omega\) in uniform magnetic field \(B_{0}\) along axis of the disc then no EMF is induced in the disc. Reason: Whenever a conductor cuts across magnetic lines of flux, an EMF is induced in the conductor. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\) Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3} .\) The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)
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