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Chapter 13: Problem 1876

A He nucleus makes a full rotation in a circle of radius \(0.8\) meter in $2 \mathrm{sec}\(. The value of the mag. field \)\mathrm{B}$ at the centre of the circle will be \(\quad\) Tesla. (a) \(\left(10^{-19} / \mu_{0}\right)\) (b) \(10^{-19} \mu_{0}\) (c) \(2 \times 10^{-10} \mathrm{H}_{0}\) (d) \(\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]\)

Short Answer

Expert verified
The short answer for the given problem is (d) \( \left[\left(2 \times 10^{-10}\right) / \mu_{0}\right] \) Tesla.

Step by step solution

01

Calculate the Magnetic Moment

First, we need to calculate the magnetic moment (µ) of the rotating He nucleus. The magnetic moment is given by the formula: \( \mu = IA \) where I is the current and A is the area enclosed by the circular path. To find I, we need to know the charge (q) and the time (t) for one complete rotation: \( I = \frac{q}{t} \) Since a He nucleus has 2 protons, its charge is twice the charge of a proton: \( q = 2e \) and t is given as 2 seconds. The area A is given by the formula for the area of a circle: \( A = \pi r^2\) Now we can calculate µ using these values. #Step 2: Calculate the Magnetic Field#
02

Calculate the Magnetic Field

Next, we need to calculate the magnetic field (B) at the center of the circle. Using the magnetic moment, the magnetic field is determined by the formula: \( B = \frac{\mu}{2 \pi r^3} \) Now substitute the calculated magnetic moment and the given radius (r = 0.8m) into this equation to find the magnetic field B. #Step 3: Match the answer choice#
03

Match the answer choice

Compare the calculated value of the magnetic field B with the answer choices given. Ensure your calculation matches the appropriate form for the correct answer choice. Using these steps, you should be able to find the correct answer to be (d) \( \left[\left(2 \times 10^{-10}\right) / \mu_{0}\right] \) Tesla.

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Most popular questions from this chapter

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 \((2 \pi \mathrm{r})\)

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

Needles \(\mathrm{N}_{1}, \mathrm{~N}_{2}\) and \(\mathrm{N}_{3}\) are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will (a) Attract \(\mathrm{N}_{1}\) strongly, \(\mathrm{N}_{2}\) weakly and repel \(\mathrm{N}_{3}\) weakly (b) Attract \(\mathrm{N}_{1}\) strongly, but repel \(\mathrm{N}_{2}\) and \(\mathrm{N}_{3}\) weakly (c) Attract all three of them (d) Attract \(\mathrm{N}_{1}\) and \(\mathrm{N}_{2}\) strongly but repel \(\mathrm{N}_{2}\)

Two concentric co-planar circular Loops of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) carry currents of respectively \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) in opposite directions. The magnetic induction at the centre of the Loops is half that due to \(\mathrm{I}_{1}\) alone at the centre. If \(\mathrm{r}_{2}=2 \mathrm{r}_{1}\) the value of $\left(\mathrm{I}_{2} / \mathrm{I}_{1}\right)$ is (a) 2 (b) \(1 / 2\) (c) \(1 / 4\) (d) 1
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