JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 8)
Two magnetic dipoles X and Y are placed at
a separation d, with their axes perpendicular to
each other. The dipole moment of Y is twice
that of X. A particle of charge q is passing,
through their midpoint P, at angle q = 45° with
the horizontal line, as shown in figure. What
would be the magnitude of force on the particle
at that instant ?
(d is much larger than the dimensions of the dipole)_8th_April_Evening_Slot_en_8_1.png)
(d is much larger than the dimensions of the dipole)
$$ \left( {{{{\mu _0}} \over {4\pi }}} \right){2M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
$$ \left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
$$\sqrt 2 \left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv$$
0
Explanation
$${\overrightarrow F _m} = q\left( {\overrightarrow V \times \overrightarrow B } \right)$$
$$\overrightarrow B = {\overrightarrow B _x} + {\overrightarrow B _y}$$_8th_April_Evening_Slot_en_8_2.png)
Since My = 2Mx
$$ \Rightarrow \left| {{{\overrightarrow B }_x}} \right| = \left| {{{\overrightarrow B }_y}} \right|$$
$${\overrightarrow B _{net}}$$ is parallel to $$\overrightarrow V $$
$$ \Rightarrow $$ $$\overrightarrow F = 0$$
$$\overrightarrow B = {\overrightarrow B _x} + {\overrightarrow B _y}$$
Since My = 2Mx$$ \Rightarrow \left| {{{\overrightarrow B }_x}} \right| = \left| {{{\overrightarrow B }_y}} \right|$$
$${\overrightarrow B _{net}}$$ is parallel to $$\overrightarrow V $$
$$ \Rightarrow $$ $$\overrightarrow F = 0$$
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