/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A long solenoid has 200 turns pe... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 76

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) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The magnetic field at the center of the second solenoid is approximately \(1.05 \times 10^{-3} \mathrm{Wb} / \mathrm{m}^{2}\), which corresponds to option (D).

Step by step solution

01

Find the current in the first solenoid.

We are given the magnetic field at the center of the first solenoid as \(6.28 \times 10^{-2} \mathrm{Wb} / \mathrm{m}^{2}\) and the number of turns per unit length as 200 turns per cm. We can use the formula for magnetic field at the center of a solenoid to find the current i flowing through it: \begin{equation} B = \mu_0 n i \end{equation} Rearrange the formula to solve for current i: \begin{equation} i = \frac{B}{\mu_0 n} \end{equation} We know that \(\mu_0 = 4\pi \times 10^{-7} \mathrm{T\cdot m/A}\). We have to convert n from turns per cm to turns per meter: \(n = 200 \times 100 = 20000 \, \text{turns/m}\). Now substitute the given values and find i: \begin{equation} i = \frac{6.28 \times 10^{-2}}{4\pi \times 10^{-7} \times 20000} \approx 1.575 \, \text{A} \end{equation}
02

Find the current in the second solenoid.

We are given that the current in the second solenoid is \(\frac{i}{3}\): \begin{equation} i_2 = \frac{i}{3} = \frac{1.575}{3} \approx 0.525\, \text{A} \end{equation}
03

Find the magnetic field in the second solenoid.

Now we have to substitute values in the formula for magnetic field at the center of a solenoid for the second solenoid. The number of turns per unit length n is 100 turns per cm which is equal to \(100 \times 100 = 10000 \, \text{turns/m}\). Substitute the values in the formula as follows: \begin{equation} B_2 = \mu_0 n_2 i_2 = 4\pi \times 10^{-7} \times 10000 \times 0.525 \approx 1.05 \times 10^{-3} \mathrm{Wb} / \mathrm{m}^{2} \end{equation} So the magnetic field at the center of the second solenoid is approximately \(1.05 \times 10^{-3} \mathrm{Wb} / \mathrm{m}^{2}\), which corresponds to option (D).

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Most popular questions from this chapter

Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(i_{1}\) and \(i_{2}\). The magnitude of the magnetic induction at a point \(P\) at a distance \(a\) from the point \(O\) in a direction perpendicular to the plane \(A C B D\) is (A) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}+i_{2}\right)\) (B) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}-i_{2}\right)\) (C) \(\frac{\mu_{0}}{2 \pi a}\left(i_{1}^{2}+i_{2}^{2}\right)^{1 / 2}\) (D) \(\frac{\mu_{0}}{2 \pi a} \frac{i_{1} i_{2}}{\left(i_{1}+i_{2}\right)}\)

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) \(\frac{\mu_{0} i}{2 \pi}\)

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 short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\), such that their centers coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{M \sqrt{5}}{d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{3 M}{d^{3}}\)
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