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Chapter 13: Problem 1886

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current of $2.5 \mathrm{Amp}$. The mag. field at its centre is tesla. (a) \(\pi \times 10^{-2}\) (b) \(2 \pi \times 10^{-2}\) (c) \(3 \pi \times 10^{-2}\) (d) \(4 \pi \times 10^{-2}\)

Short Answer

Expert verified
The magnetic field at the center of the solenoid is (a) \(\pi \times 10^{-2} \mathrm{T}\).

Step by step solution

01

Convert turns per centimeter to turns per meter

\ First, we will need to convert the given number of turns per centimeter to turns per meter. Recall that there are 100 centimeters in 1 meter. So, \(n = 200 \text{ turns/cm} \times 100 \text{ cm/m} = 20000 \text{ turns/m}\).
02

Recall the formula for the magnetic field inside a long solenoid

\ The formula for the magnetic field inside a long solenoid is \(B = \mu_0 n I\), where \(B\) is the magnetic field, \(\mu_0\) is the vacuum permeability constant (\(4\pi \times 10^{-7} \text{ T m/A}\)), \(n\) is the number of turns per meter, and \(I\) is the current.
03

Plug in the values and solve for the magnetic field

\ Now, plug the given values into the formula and solve for the magnetic field: \(B = (4\pi \times 10^{-7} \text{ T m/A}) \times (20000 \text{ turns/m}) \times (2.5 \mathrm{A})\).
04

Calculate the magnetic field

\ Using a calculator, compute the magnetic field: \(B = (4\pi \times 10^{-7}) \times (20000) \times (2.5)\) \(B = \pi \times 10^{-2} \mathrm{T}\).
05

Compare the result to the given options

\ Our calculated magnetic field is \(\pi \times 10^{-2} \text{ T}\), which matches option (a). Therefore, the correct answer is (a) \(\pi \times 10^{-2} \text{ T}\).

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

The true value of angle of dip at a place is \(60^{\circ}\), the apparent dip in a plane inclined at an angle of \(30^{\circ}\) with magnetic meridian is. (a) \(\tan ^{-1}(1 / 2)\) (b) \(\tan ^{-1} 2\) (c) \(\tan ^{-1}(2 / 3)\) (d) None of these

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 particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)
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