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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

A thin magnetic needle oscillates in a horizontal plane with a period \(\mathrm{T}\). It is broken into n equal parts. The time period of each part will be (a) \(\mathrm{T}\) (b) \(\mathrm{n}^{2} \mathrm{~T}\) (c) \((\mathrm{T} / \mathrm{n})\) (d) \(\left(\mathrm{T} / \mathrm{n}^{2}\right)\)

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other. 3 Amp and \(4 \mathrm{Amp}\) are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be Tesla. (a) \(5 \times 10^{-5}\) (b) \(7 \times 10^{-5}\) (c) \(12 \times 10^{-5}\) (d) \(10^{-5}\)

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) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)
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