JEE PYQ: Electrostatics Question 40
Question 40 - 2019 (08 Apr Shift 2)
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)
(1) $\left(\frac{\mu_0}{4\pi}\right)\frac{M}{\left(\frac{d}{2}\right)^3} \times qv$
(2) 0
(3) $\sqrt{2}\left(\frac{\mu_0}{4\pi}\right)\frac{M}{\left(\frac{d}{2}\right)^3} \times qv$
(4) $\left(\frac{\mu_0}{4\pi}\right)\frac{2M}{\left(\frac{d}{2}\right)^3} \times qv$
Show Answer
Answer: (2)
Solution
At the midpoint P, the field from dipole X is $B_1 = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$ and from dipole Y is $B_2 = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3}$. Since the axes are perpendicular and the resultant field at $\theta = 45°$ is at the equatorial/axial configuration, $\tan\theta = \frac{B_2}{B_1} = 1$, giving $\theta = 45°$. The force $F = qv \times B$ where $\vec{v} \parallel \vec{B}$, so $F = 0$.
BETA
