AP EAMCET · PHYSICS · Magnetic Effects of Current
A particle of charge \(1.0 \times 10^{-16} .{ }^6 \mathrm{C}\) moves through a uniform magnetic field \(\mathbf{B}=B_0(\hat{i}+4 \hat{j}) \mathrm{T}\). The particle velocity at some instant is \(\mathbf{v}=(2 \hat{i}+4 \hat{j}) \mathrm{ms}^{-1}\) and the magnetic force acting on it is \(3 \times 10^{-16} \hat{k} \mathrm{~N}\). The magnitude of \(B_0\) is
- A \(1.0 \mathrm{~T}\)
- B \(2.5 \mathrm{~T}\)
- C \(0.5 \mathrm{~T}\)
- D \(0.75 \mathrm{~T}\)
Answer & Solution
Correct Answer
(D) \(0.75 \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Force on a charged particle moving in region of a magnetic field is given by \(F=q(\mathbf{v} \times \mathbf{B})\) ...(i) Here, \(\quad q=1.0 \times 10^{-16} \mathrm{C}\)…
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