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Chapter 13: Problem 1901

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

Short Answer

Expert verified
The magnetic induction at the axis inside the solenoid is \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

Step by step solution

01

Identify the given parameters

We are given the length of the solenoid \(l\), diameter \(d\), number of turns per centimeter \(n\), and the current flowing through the solenoid \(I\). These values are: Length of solenoid (\(l\)) = \(1.5\ \mathrm{m}\) Diameter of solenoid (\(d\)) = \(4\ \mathrm{cm}\) Turns per centimeter (\(n\)) = \(10\ \mathrm{turns/cm}\) Current through solenoid (\(I\)) = \(5\ \mathrm{A}\)
02

Calculate the total number of turns in the solenoid

To find the total number of turns in the solenoid, we must multiply the number of turns per centimeter by the length of the solenoid in centimeters: Total number of turns (\(N\)) = \(n \times l\) We have \(l = 1.5\ \mathrm{m} = 150\ \mathrm{cm}\) (converting meters to centimeters). So: \(N = 10\ \mathrm{turns/cm} \times 150\ \mathrm{cm} = 1500\ \mathrm{turns}\)
03

Apply Ampere's Law to find magnetic field (induction) B

Ampere's law states that the magnetic field (induction) inside a solenoid can be calculated as: \(B = \mu_0 n I \) Here, \(\mu_0\) is the permeability of free space, and its value is \( 4\pi \times 10^{-7}\ \mathrm{Tm/A} \) \(n\) is the density of turns (turns per meter) and requires conversion from turns per centimeter. Density of turns in \(\mathrm{turns/m}\) will be: \( n = 10\ \mathrm{turns/cm} \cdot (\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}) = 1000\ \mathrm{turns/m} \) Now we can calculate B: \( B = (4\pi \times 10^{-7}\ \mathrm{Tm/A})\times (1000\ \mathrm{turns/m})\times(5\ \mathrm{A}) \)
04

Evaluate the magnetic induction at the axis inside the solenoid

Now, we can evaluate B: \(B = 4\pi \times 10^{-4}\ \mathrm{T}\) This result matches the option (a) \(2\pi \times 10^{-3}\ \mathrm{T}\), so the magnetic induction at the axis inside the solenoid is: \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

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Most popular questions from this chapter

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. A horizontal magnetic field \(\mathrm{B}=6 \times 10^{-4}\) Tesla exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\)

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

If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be (a) Only inside the pipe (b) Only outside the pipe (c) Neither inside nor outside the pipe (d) Both inside and outside the pipe

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.
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