Question

(a) A particle of charge q is moving with velocity v in the presence of crossed electric field E and magnetic field B as shown. Write the condition under which the particle will continue moving along x-axis. How would the trajectory of the particle be affected if the electric field is switched off?

(b) A horizontal wire AB of length 'l' and mass ‘m’ carries a steady current I₁, free to move in vertical plane is in equilibrium at a height of 'h’ over another parallel long wire CD carrying a steady current I₂, which is fixed in a horizontal plane as shown. Derive the expression for the force acting per unit length on the wire AB and write the condition for which wire AB is in equilibrium.

Answer 1

(a) The condition for particle to continue along x-axis is:

qE = q vB sin 90°

or v = \(\frac{E}{B}\) 

When the electric field is switched off, the trajectory of the charge will become helical about the direction of the magnetic field.

(b) At any point P magnetic field intensity due to current I₂ in CD

B = \(\frac{\mu_0I_2}{2\pi r}\)

Its direction is perpendicular to the plane of the paper and downwards.

Force per unit length experienced by conductor AB is given by

\(\frac{F}{h} \) = B, I₁ = \(\frac{\mu_0I_1I_2}{2\pi r}\)

At equilibrium

\(\frac{\mu_0I_1I_2}{2\pi r} = \frac{mg}{h}\)