JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 21)
A long, straight wire of radius a carries a current
distributed uniformly over its cross-section. The
ratio of the magnetic fields due to the wire at
distance
$${a \over 3}$$
and 2$$a$$, respectively from the axis
of the wire is :
2
$${1 \over 2}$$
$${3 \over 2}$$
$${2 \over 3}$$
Explanation
_9th_January_Morning_Slot_en_21_1.png)
Let current density be J.
$$ \therefore $$ Applying Ampere's law. For point P
$$\oint {\overrightarrow {B.} d\overrightarrow l } = {\mu _0}i$$
$$ \Rightarrow $$ BP2$$\pi $$$${a \over 3}$$ = $${\mu _0}J\pi {{{a^2}} \over 9}$$ ...(1)
$$ \therefore $$ Applying Ampere's law. For point Q
BQ2$$\pi $$$${(2a)}$$ = $${\mu _0}J\pi {a^2}$$ ...(2)
Dividing (1) by (2) we get
$${{{B_P}2\pi {a \over 3}} \over {{B_Q}4\pi a}} = {{{\mu _0}J\pi {{{a^2}} \over 9}} \over {{\mu _0}J\pi {a^2}}}$$
$$ \Rightarrow $$ $${{{B_P}} \over {{B_Q}}} = {2 \over 3}$$
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