JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 14)
A Helmholtz coil has a pair of loops, each with $$N$$ turns and radius $$R$$. They are placed coaxially at distance $$R$$ and the same current $${\rm I}$$ flows through the loops in the same direction. $$P,$$ midway between the centers $$A$$ and $$C$$, is given by [Refer to figure given below] :
_15th_April_Morning_Slot_en_14_1.png)
$${{8N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}$$
$${{8N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}$$
$${{4N{\mu _0}{\rm I}} \over {{5^{1/2}}R}}$$
$${{4N{\mu _0}{\rm I}} \over {{5^{3/2}}R}}$$
Explanation
P is the mid-point of line AC. A and C are the center of the two circle of each radius R.
Current flows through loop A and B are in same direction, So the magnetic field will also be in the same direction. Magnitude of magnetic field at paint P
= magnitude of magnetic field due to A and B at paint P.
= 2 $$\left[ {{{{\mu _0}NI\,{R^2}} \over {2{{\left( {{R^2} + {{{R^2}} \over 4}} \right)}^{{3 \over 2}}}}}} \right]$$
$$=$$ $${{{\mu _0}\,NI{R^2}} \over {{{{5^{{3 \over 2}}}{R^3}} \over 8}}}$$
= $${{8\,{\mu _0}NI} \over {{5^{{3 \over 2}}}\,R}}$$
Current flows through loop A and B are in same direction, So the magnetic field will also be in the same direction. Magnitude of magnetic field at paint P
= magnitude of magnetic field due to A and B at paint P.
= 2 $$\left[ {{{{\mu _0}NI\,{R^2}} \over {2{{\left( {{R^2} + {{{R^2}} \over 4}} \right)}^{{3 \over 2}}}}}} \right]$$
$$=$$ $${{{\mu _0}\,NI{R^2}} \over {{{{5^{{3 \over 2}}}{R^3}} \over 8}}}$$
= $${{8\,{\mu _0}NI} \over {{5^{{3 \over 2}}}\,R}}$$
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