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Chapter 17: Problem 3

The value of magnetic field between plates of capacitor, at distance of \(1 \mathrm{~m}\) from centre where electric field varies by \(10^{10} \mathrm{~V} / \mathrm{m} / \mathrm{s}\) will be (A) \(5.56 \mathrm{~T}\) (B) \(5.56 \mu \mathrm{T}\) (C) \(5.56 \mathrm{mT}\) (D) \(55.6 \mathrm{nT}\)

Short Answer

Expert verified
(C) 33.33 mT

Step by step solution

01

Identify the given values

The problem gives us the rate of change of the value of the electric field (E') which is \(10^{10}\) V/m/s.
02

Invoke Faraday's law and solve for B

Using the formula B = E / c, where c is the speed of light (\(3 \times 10^{8}\) m/s), and given that E changes by E' per second, we substitute E' for E in the formula to find B. So, B = \(E' / c = 10^{10} / 3 \times 10^{8} = 0.033\) T or 33.33 mT.

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