WBJEE · Physics · Magnetic Effects of Current
A charged particle moving with a volocity \(\vec{v}=v_1 \hat{i}+v_2 \hat{j}\) in a magnetic field \(\vec{B}\) experiences a force \(\vec{F}=F_1 \hat{i}+F_2 \hat{j}\). Here \(v_1 \cdot v_2, F_1, F_2\) all are constants. Then \(\vec{B}\) can be
- A \(\overrightarrow{\mathrm{B}}=\mathrm{B}_1 \hat{\mathrm{I}}+\mathrm{B}_2 \hat{\mathrm{j}}\) with \(\frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{B}_1}{\mathrm{~B}_2}\)
- B \(\vec{B}=B_1 \hat{i}+B_2 \hat{j}+B_3 \hat{k}\) with \(\frac{v_1}{v_2}=\frac{B_1}{B_2}\)
- C \(\vec{B}=B_3 \hat{j}\) with \(B_1=B_2=0\)
- D \(\overrightarrow{\mathrm{B}}=\mathrm{B}_1 \hat{\mathrm{j}}+\mathrm{B}_2 \hat{\mathrm{k}}\) with \(\frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{v}_1}{\mathrm{v}_2}\)
Answer & Solution
Correct Answer
(B) \(\vec{B}=B_1 \hat{i}+B_2 \hat{j}+B_3 \hat{k}\) with \(\frac{v_1}{v_2}=\frac{B_1}{B_2}\)
Step-by-step Solution
Detailed explanation
Hint: \(\vec{F} \cdot \vec{V}=0 \Rightarrow \frac{F_1}{F_2}=-\frac{V_2}{V_1} \ldots \ldots . .(I)\)…
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