JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 9)
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d > > a). If the loop a applies a force F on the wire then :
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$$F = 0$$
$$F \propto \left( {{a \over d}} \right)$$
$$F \propto \left( {{{{a^2}} \over {{d^3}}}} \right)$$
$$F \propto {\left( {{a \over d}} \right)^2}$$
Explanation
Magnetic field due to current carrying loop at distance d, is,
B = $${{{\mu _0}} \over {4\pi }} \times {{\overrightarrow M \times \overrightarrow d } \over {{d^3}}}$$
= $${{{\mu _0}} \over {4\pi }} \times {{i \times \pi {a^2} \times d \times \sin {{90}^o}} \over {{d^3}}}$$ [as M = iA]
= $${{{\mu _0}} \over {4\pi }} \times {{i\pi {a^2}} \over {{d^2}}}$$
$$ \therefore $$ B $$ \propto $$ $${{{a^2}} \over {{d^2}}}$$
We also know,
F = Bil
$$ \therefore $$ F $$ \propto $$ B
So, F $$ \propto $$ $${{{a^2}} \over {{d^2}}}$$
B = $${{{\mu _0}} \over {4\pi }} \times {{\overrightarrow M \times \overrightarrow d } \over {{d^3}}}$$
= $${{{\mu _0}} \over {4\pi }} \times {{i \times \pi {a^2} \times d \times \sin {{90}^o}} \over {{d^3}}}$$ [as M = iA]
= $${{{\mu _0}} \over {4\pi }} \times {{i\pi {a^2}} \over {{d^2}}}$$
$$ \therefore $$ B $$ \propto $$ $${{{a^2}} \over {{d^2}}}$$
We also know,
F = Bil
$$ \therefore $$ F $$ \propto $$ B
So, F $$ \propto $$ $${{{a^2}} \over {{d^2}}}$$
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