WBJEE · Physics · Magnetic Effects of Current
A straight wire is placed in a magnetic field that varies with distance \(x\) from origin as \(\vec{B}=B_0\left(2-\frac{x}{a}\right) \hat{k}\). Ends of wire are at \((a, 0)\) and \((2 a, 0)\) and it carries a current \(\mathrm{I}\). If force on wire is \(\vec{F}=\mathrm{IB}_0\left(\frac{k a}{2}\right) \hat{\mathrm{j}}\), then value of \(k\) is
- A 1
- B 5
- C \(-1\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(-1\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{dF}}=\mathrm{Idx} \times \overrightarrow{\mathrm{B}}=\operatorname{Idx} \hat{\mathrm{i}} \times \mathrm{B} \hat{\mathrm{k}}=\operatorname{IdxB}(-\hat{\mathrm{j}})\)…
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