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Chapter 14: Problem 2040

A conducting rod \(P Q\) of length \(4 \ell\) is rotated about a point \(O\) in a uniform magnetic field \(\mathrm{B} \rightarrow \mathrm{PO}=\ell\) Then (a) $\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}=-\left[\left\\{\mathrm{B} \omega \ell^{2}\right\\} /\right.$ \(\\{2\\}] \quad\) (b) $\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{O}}=(5 / 2) \mathrm{B} \operatorname{co} \ell^{2}$ (c) $\mathrm{V}_{Q}-\mathrm{V}_{\mathrm{O}}=(9 / 2) \mathrm{B} \omega \ell^{2}$ (d) $\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}=4 \mathrm{~B} \omega \ell^{2}$

Short Answer

Expert verified
The short answer is: (c) V_Q - V_O = (9 / 2) Bωℓ^2.

Step by step solution

01

Find the tangential velocities

To find the tangential velocities at points P and Q, we'll use the formula: v = ωr where ω is the angular velocity of the rod about point O, and r is the distance from point O to the point of interest (P or Q). For point P: u = ω(ℓ) For point Q: v = ω(5ℓ)
02

Calculate the emf induced in the rod

Using the formula given for the emf induced in a rotating rod in a magnetic field: emf = Bℓ(v - u) We substitute the tangential velocities found in step 1: emf = Bℓ(ω(5ℓ) - ω(ℓ)) emf = Bℓω(5ℓ - ℓ) emf = 4Bℓ^2ω
03

Determine the potential difference between points P and Q, O and Q

The potential difference between points P and Q is the emf: V_Q - V_P = 4Bℓ^2ω Now, let's find the potential difference between points O and Q: V_Q - V_O = V_Q - (V_P + 4Bℓ^2ω) Using the relation found above: V_Q - V_P = 4Bℓ^2ω, V_Q - V_O = 9Bℓ^2ω / 2
04

Compare the results with given options

Comparing the results from steps 3 with the given options, we can see that: (a) is incorrect because the sign and fraction are not correct. (b) is incorrect because co is not part of the expression. (c) is correct as it matches the expression for the potential difference between points Q and O derived in our solution. (d) is incorrect because the expression is for potential difference between points P and Q and the factor is not correct. So, the correct answer is (c) V_Q - V_O = (9 / 2) Bωℓ^2.

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Most popular questions from this chapter

A coil has an area of \(0.05 \mathrm{~m}^{2}\) \& it has 800 turns. It is placed perpendicularly in a magnetic field of strength, $4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{-2}\(. It is rotated through \)90^{\circ}\( in \)0.1 \mathrm{sec}$. The average emf induced in the coil is... (a) \(0.056 \mathrm{v}\) (b) \(0.046 \mathrm{v}\) (c) \(0.026 \mathrm{v}\) (d) \(0.016 \mathrm{v}\)

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