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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A particle of charge \(16\times10^{-16}\, C\) moving with velocity \(10\, ms^{-1}\) along \(x-\) axis enters a region where magnetic field of induction \(\vec B\) is along the \(y-\) axis and an electric field of magnitude \(10^4\, Vm^{-1}\) is along the negative \(z-\) axis. If the charged particle continues moving along \(x-\) axis, the magnitude of \(\vec B\) is
- A \(16\times10^3\, Wb \,m^{ -2}\)
- B \(2\times10^3\, Wb \,m^{ -2}\)
- C \(1\times10^3\, Wb \,m^{ -2}\)
- D \(4\times10^3\, Wb \,m^{ -2}\)
Answer & Solution
Correct Answer
(C) \(1\times10^3\, Wb \,m^{ -2}\)
Step-by-step Solution
Detailed explanation
Since particle is moving undeflected So, \(qE\, = qvB\) \( \Rightarrow B = \frac{E}{V} = \frac{{{{10}^4}}}{{10}} = {10^3}\,wb/{m^2}\)
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