JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 13)
Two long parallel conductors S1 and S2 are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).
A charge particle of 3$$\pi$$ coulomb is passing through the point P with velocity $$\overrightarrow v = (2\widehat i + 3\widehat j)$$ m/s; where $$\widehat i$$ and $$\widehat j$$ represents unit vector along x & y axis respectively.
The force acting on the charge particle is $$4\pi \times {10^{ - 5}}( - x\widehat i + 2\widehat j)$$ N. The value of x is :
_27th_June_Evening_Shift_en_13_1.png)
Explanation
Field at P is $$ = \left( {{{{\mu _0} \times {i_1}} \over {2\pi {r_1}}} - {{{\mu _0}{i_2}} \over {2\pi {r_2}}}} \right)\left( { - \widehat k} \right)$$
$$ = - \left( {{{{\mu _0}4} \over {2\pi \times 0.04}} - {{{\mu _0} \times 2} \over {2\pi \times 0.06}}} \right)\widehat k = - {{{\mu _0} \times 200} \over {6\pi }}\widehat k$$
So, force $$\overrightarrow F = q\overrightarrow v \times \overrightarrow B $$
$$ = 3\pi \left( {2\widehat i + 3\widehat j} \right) \times \left( { - \left( {{{{\mu _0} \times 200} \over {6\pi }}\widehat k} \right)} \right)$$
$$ = 3\pi \left( {{{200{\mu _0}} \over {3\pi }}\widehat j - {{100{\mu _0}} \over \pi }\widehat i} \right)$$
$$ = 200{\mu _0}\widehat j - 300{\mu _0}\widehat i$$
$$ = 4\pi \times {10^{ - 5}}(2\widehat j - 3\widehat i)$$
So, $$x = 3$$
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