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 θ = 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) (a) ((μ₀)/(4π))2M(d/2)³ × qv (b) ((μ₀)/(4π))M(d/2)³ × qv (c) √(2)((μ₀)/(4π))M(d/2)³ × qv (d) 0

Answer: (d). PB_XB_YB, v45° P lies on the axis of X: BX = (μ₀)/(4π)(2M)/((d/2)³), along X's moment (S to N), i.e. to the right. P lies on the equatorial line of Y: BY = (μ₀)/(4π)(2M)/((d/2)³), opposite to Y's moment (Y points down), i.e. upward. BX = BY, so the resultant B points at 45° to the horizontal. The velocity is also at 45°, so v ∥ B and F = qvBsin 0° = 0. So the answer is (d) 0.

Two magnetic dipoles X and Y are placed at a separation d, with their… | JEE Main 2019 (8 Apr, Shift 2)

Q16JEE Main 2019 · 8 APR · SHIFT 2Difficult

Field of a Magnet & Magnet in a Uniform Field

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 $\theta$ = 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)

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Answer: (d)

P lies on the axis of X: $B_X = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$ , along X's moment (S to N), i.e. to the right.
P lies on the equatorial line of Y: $B_Y = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$ , opposite to Y's moment (Y points down), i.e. upward.
$B_X = B_Y$ , so the resultant $\vec{B}$ points at $45^\circ$ to the horizontal.
The velocity is also at $45^\circ$ , so $\vec{v} \parallel \vec{B}$ and $F = qvB\sin 0^\circ = 0$ .
So the answer is (d) 0.

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Field of a Magnet & Magnet in a Uniform Field question from JEE Main 2019 (8 Apr, Shift 2)

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