JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 27)
A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to :
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1.0 $$ \times $$ 10$$-$$7 T
1.5 $$ \times $$ 10$$-$$7 T
1.5 $$ \times $$ 10$$-$$5 T
1.0 $$ \times $$ 10$$-$$5 T
Explanation
Magnetic field due to circular arc
$$\overrightarrow B = {{{\mu _0}\,i\,\theta } \over {4\pi r}}$$
Due to QR arc magnetic field is outward direction of the plane.
And due to PS arc magnetic field is inward direction of the plane,
So, net magnetic field,
$$\overrightarrow B = \left( {{{\overrightarrow B }_{QR}} - {{\overrightarrow B }_{PS}}} \right)\widehat K$$
$$\overrightarrow B = {{{\mu _0}i\theta } \over {4\pi }}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$$
$$\overrightarrow B = {10^{ - 7}} \times 10 \times {\pi \over 4}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$$
$$\left| {\overrightarrow B } \right| = 1 \times {10^{ - 5}}\,\,T$$
$$\overrightarrow B = {{{\mu _0}\,i\,\theta } \over {4\pi r}}$$
Due to QR arc magnetic field is outward direction of the plane.
And due to PS arc magnetic field is inward direction of the plane,
So, net magnetic field,
$$\overrightarrow B = \left( {{{\overrightarrow B }_{QR}} - {{\overrightarrow B }_{PS}}} \right)\widehat K$$
$$\overrightarrow B = {{{\mu _0}i\theta } \over {4\pi }}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$$
$$\overrightarrow B = {10^{ - 7}} \times 10 \times {\pi \over 4}\left( {{1 \over {3 \times {{10}^{ - 2}}}} - {1 \over {5 \times {{10}^{ - 2}}}}} \right)\widehat K$$
$$\left| {\overrightarrow B } \right| = 1 \times {10^{ - 5}}\,\,T$$
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