A parabolic section of wire, as shown in figure is located in the X-Y plane and carries a current of 12A. A uniform magnetic field B = 0.4T making an angle of 600 with X-axis exists throughout the plane. The total force acting on a wire between the origin and the point x = 0.25m, y = 1.00 m.

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$- 0 .68 \hat{K} N$

$- 1.36 \hat{K} N$

$0 .68 \hat{K} N$

$1.36 \hat{K} N$

answer is D.

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Detailed Solution

$\begin{array}{l} \bar{l} = 0 .25 i + 1 j \\ \bar{B} = Bcos 60^{0} \hat{i} + Bsin 60^{0} \hat{j} \\ \bar{B} = 0 .4 \times \frac{1}{2} \hat{i} + 0 .4 \times \frac{\sqrt{3}}{2} \hat{j} \\ \bar{B} = 0 .2 \hat{i} + 0 .2 \sqrt{3} \hat{j}; \bar{F} = i (\bar{l} \times \bar{B}) \end{array}$

$\begin{array}{l} F = 12 |\begin{array}{ccl} i & j & k \\ \frac{1}{4} & 1 & 0 \\ 0 .2 & 0 .2 \sqrt{3} & 0 \end{array}| \\ F = 12 [1 .732 \times 0 .05 - 0 .2] \hat{k}; F = - 1 .36 \hat{k} \end{array}$