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Chapter 15: Problem 2113

The dimensional formula of \(\mu_{0} \mathrm{E}_{0}\) is (A) \(L^{2} T^{-2}\) (B) \(L^{-2} T^{2}\) (C) \(\mathrm{L}^{1} \mathrm{~T}^{-1}\) (D) \({L}^{-1} \mathrm{~T}^{1}\)

Short Answer

Expert verified
The dimensional formula of \(\mu_0 E_0\) is \(M L^3 T^{-2} A^{-3}\), which does not match any of the given options. There must be an error in the given options, so the correct answer is not presented in the options provided.

Step by step solution

01

Find the dimensions of \(\mu_0\)

The permeability constant (\(\mu_0\)) has dimensions of \(T^2 A^{-2}\).
02

Find the dimensions of \(E_0\)

The electric field constant (\(E_0\)) has dimensions of \(M L^3 T^{-4} A^{-1}\). #Step 2: Calculate the dimensions of the product#
03

Calculate the dimensions of \(\mu_0 E_0\)

Multiply the dimensions of \(\mu_0\) with the dimensions of \(E_0\) to find the dimensions of the product: \[(T^2 A^{-2}) \times (M L^3 T^{-4} A^{-1}) = M L^3 T^{-2} A^{-3}\] #Step 3: Compare with the given options#
04

Analyze the dimensions

None of the given options match the dimensions we calculated in the previous step. This means that there must be an error in the given options, and the correct answer is not presented in the options provided.

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Most popular questions from this chapter

The rms value of the electric field of the light coming from the sun is $720 \mathrm{~N} / \mathrm{c}$. The average total energy density of the electromagnetic wave is (A) \(4.58 \times 10^{-6} \mathrm{Jm}^{-3}\) (B) \(6.3 \times 10^{-9} \mathrm{Jm}^{-3}\) (C) \(81.35 \times 10^{-12} \mathrm{Jm}^{-3}\) (D) \(3.3 \times 10^{-3} \mathrm{Jm}^{-3}\)

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