AP EAMCET · PHYSICS · Magnetic Effects of Current
A velocity selector is to be constructed to select ions with a velocity of \(6 \mathrm{~km} \mathrm{~s}^{-1}\).
If the electric field used is \(400 \mathrm{~V} \mathrm{~m}^{-1}\), then the magnetic field to be used is
- A \(\frac{11}{20} \mathrm{~T}\)
- B \(\frac{2}{3} \mathrm{~T}\)
- C \(\frac{1}{15} \mathrm{~T}\)
- D \(\frac{2}{15} \mathrm{~T}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{15} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
\(B = \frac{E}{v}\) \(B = \frac{400 \mathrm{~V} \mathrm{~m}^{-1}}{6 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}}\) \(B = \frac{1}{15} \mathrm{~T}\)
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