JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A conductor lies along the \(z-\)axis \(a\) \(-1.5\)\( \le Z < 1.5\,m\) carries a fixed current of \(10.0\ A\) in \( - {\hat a_z}\) direction (see figure). For a field \(\vec B\) \(=\) \(3.0 \times 10^{-4}\) \(e^{-0.2x}\) \({\hat a_y}\,T\) find the power required to move the conductor at constant speed to \(x = 2.0\ m, y = 0\ m\) in \(5 \times 10^{-3}\ s\). Assume parallel motion along the \(x-\)axis........\( W\)

- A \(2.97\)
- B \(14.85\)
- C \(29.7 \)
- D \(1.57\)
Answer & Solution
Correct Answer
(A) \(2.97\)
Step-by-step Solution
Detailed explanation
Work done in moving the conductor is, \(W=\int_{0}^{2} F d x\) \(=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x\) \(=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x\)…
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