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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 magnet of magnetic moment \(\mathrm{M}\) and pole strength \(\mathrm{m}\) is divided in two equal parts, then magnetic moment of each part will be (a) \(\mathrm{M}\) (b) \((\mathrm{M} / 2)\) (c) \((\mathrm{M} / 4)\) (d) \(2 \mathrm{M}\)

The coercively of a bar magnet is \(100 \mathrm{~A} / \mathrm{m}\). It is to be diamagnetism by placing it inside a solenoid of length \(100 \mathrm{~cm}\) and number of turns 50 . The current flowing through the solenoid will be (a) \(4 \mathrm{~A}\) (b) \(2 \mathrm{~A}\) (c) \(1 \mathrm{~A}\) (d) Zero

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A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that \(\mathrm{B}^{-}\) is in plane of the coil. If due to a current \(\mathrm{I}\) in the triangle a torque \(\tau\) acts on it, the side 1 of the triangle is (a) \((2 / \sqrt{3})(\tau / \mathrm{BI})^{1 / 2}\) (b) \((2 / 3)(\tau / B I)\) (c) \(2[\tau /\\{\sqrt{(} 3) \mathrm{BI}\\}]^{1 / 2}\) (d) \((1 / \sqrt{3})(\tau / \mathrm{BI})\)

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