KCET · Physics · Magnetic Effects of Current
A charged particle with a velocity \(2 \times 10^{3} \mathrm{~ms}^{-1}\) passes undeflected through electric field and magnetic fields in mutually perpendicular directions. The magnetic field is \(1.5 \mathrm{~T}\). The magnitude of electric field will be
- A \(1.5 \times 10^{3} \mathrm{NC}^{-1}\)
- B \(2 \times 10^{3} \mathrm{NC}^{-1}\)
- C \(3 \times 10^{3} \mathrm{NC}^{-1}\)
- D \(1.33 \times 10^{3} \mathrm{NC}^{-1}\)
Answer & Solution
Correct Answer
(C) \(3 \times 10^{3} \mathrm{NC}^{-1}\)
Step-by-step Solution
Detailed explanation
The charged particle goes undeflected through both the fields, therefore, force experience by charged particle due magnetic field must be equal to the force experienced by the charge particle due to electric field, i.e., \(F_{m}=F_{e}\)
or \(\quad e v B \sin \theta=e E\)
Given, \(\quad v=2 \times 10^{3} \mathrm{~ms}^{-1}\)
\(\text {and } B =1.5 \mathrm{~T} \)
\( \text {Hence, } \theta =90^{\circ} \)
\( =v B \sin \theta=2 \times 10^{3} \times 1.5 \times \sin 90^{\circ} \)
\( =3 \times 10^{3} \mathrm{v} / \mathrm{m}\)
or \(\quad e v B \sin \theta=e E\)
Given, \(\quad v=2 \times 10^{3} \mathrm{~ms}^{-1}\)
\(\text {and } B =1.5 \mathrm{~T} \)
\( \text {Hence, } \theta =90^{\circ} \)
\( =v B \sin \theta=2 \times 10^{3} \times 1.5 \times \sin 90^{\circ} \)
\( =3 \times 10^{3} \mathrm{v} / \mathrm{m}\)
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