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QJEEM19Ph9d11: JEE Main 2019 Physics - Class 12 JEEM19Ph9d11 Solution: Two magnetic dipoles X and Y are placed at a separation d, with …

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) JEE Main 2019 (Online) 8th April Evening Slot Physics - Magnetic Properties of Matter Question 47 English

Multiple Choice Options

  • A (μ04π)2M(d/2)3×qv\left( {{{{\mu _0}} \over {4\pi }}} \right){2M \over {{{\left( {d/2} \right)}^3}}} \times qv
  • B (μ04π)M(d/2)3×qv\left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv
  • C 2(μ04π)M(d/2)3×qv\sqrt 2 \left( {{{{\mu _0}} \over {4\pi }}} \right){M \over {{{\left( {d/2} \right)}^3}}} \times qv
  • D 0

Step-by-Step Verified Solution

F→m=q(V→×B→){\overrightarrow F _m} = q\left( {\overrightarrow V \times \overrightarrow B } \right)

B→=B→x+B→y\overrightarrow B = {\overrightarrow B _x} + {\overrightarrow B _y} JEE Main 2019 (Online) 8th April Evening Slot Physics - Magnetic Properties of Matter Question 47 English Explanation Since My = 2Mx

⇒∣B→x∣=∣B→y∣\Rightarrow \left| {{{\overrightarrow B }_x}} \right| = \left| {{{\overrightarrow B }_y}} \right|

B→net{\overrightarrow B _{net}} is parallel to V→\overrightarrow V

⇒\Rightarrow F→=0\overrightarrow F = 0

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