TS EAMCET · Physics · Magnetic Effects of Current
In Thomson's experiment to determine \(\frac{e}{m}\) of an electron, it is found that an electron beam having a kinetic energy of \(45.5 \mathrm{eV}\) remains undeflected, when subjected to crossed electric and magnetic fields. If \(E=1 \times 10^3 \mathrm{Vm}^{-1}\), the value of \(B\) is (mass of the electron is \(9.1 \times 10^{-31} \mathrm{~kg}\) )
- A \(2.5 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}\)
- B \(5.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
- C \(2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
- D \(1.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
Answer & Solution
Correct Answer
(C) \(2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
Step-by-step Solution
Detailed explanation
The kinetic energy \(K=\frac{1}{2} m v^2\) \(v^2=\frac{2 K}{m}\) \(v^2=\frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\) \(v^2=16 \times 10^{12}\) \(v=4 \times 10^6\) Again velocity, \(v=\frac{E}{B}\) \(4 \times 10^6=\frac{1 \times 10^3}{B}\)…
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