Question

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 \(\theta = 45^\circ\) 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. \(\left(\frac{\mu_0}{4\pi}\right)\frac{2M}{(d/2)^3} \times qv\)
• B. \(\left(\frac{\mu_0}{4\pi}\right)\frac{M}{(d/2)^3} \times qv\)
• C. \(\sqrt{2}\left(\frac{\mu_0}{4\pi}\right)\frac{M}{(d/2)^3} \times qv\)
• D. 0

Hint:

P lies on the axial line of X and the equatorial line of Y; find both fields, add them as vectors and compare the direction of the resultant with the velocity.

Answer 1

The Correct Option is D

Let \(k = \frac{\mu_0}{4\pi}\frac{M}{(d/2)^3}\). P is at distance \(d/2\) from each dipole.
Dipole X (moment \(M\), pointing S to N, i.e. to the right) - P is on its axial line: \(B_X = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3} = 2k\), directed along the moment, i.e. horizontally to the right.
Dipole Y (moment \(2M\), pointing S to N, i.e. downward) - P is on its equatorial line: \(B_Y = \frac{\mu_0}{4\pi}\frac{2M}{(d/2)^3} = 2k\), directed opposite to the moment, i.e. vertically upward.
Net field \(= 2k(\hat{x} + \hat{y})\): magnitude \(2\sqrt2 k\), at \(45^\circ\) above the horizontal.
The velocity is also at \(45^\circ\) above the horizontal, i.e. parallel to \(\vec{B}\). Hence \(\vec{F} = q\vec{v}\times\vec{B} = 0\).
Answer: (4) 0