A conducting rod PQ of mass m and length l is placed on two long parallel (smooth and conducting) rails connected to a capacitor as shown. The rod PQ is connected to a non conducting spring of spring constant k, which is initially in relaxed state. The entire arrangement is placed in a magnetic field perpendicular to the plane of figure. Neglect the resistance of the rails and rod. Now, the rod is imparted a velocity v₀ towards right, then acceleration of the rod as a function of its displacement x is given by

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{kx}{m - B^{2} /^{2} C}$

$\frac{kx}{m + B^{2} I^{2} C}$

$\frac{kx}{B^{2} I^{2} c}$

$\frac{kx}{m}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let the velocity of rod be v when it has been displaced by x . Due to motion of rod an emf will be induced in rod given by e = Bvl , due to this induced emf, charging of the capacitor takes place as a current, flows in the circuit [for very small time] a result of this current, the rod experiences a magnetic force given by IBl .

From Newton’s second law,
$\begin{array}{l} ∣ BI + kx = ma [a = \frac{dv}{dt}] \\ I = \frac{d}{dt} [Q] = \frac{d}{dt} [C \times BvI] = CBI \times \frac{dv}{dt} \\ \Rightarrow a = \frac{kx}{m - B^{2} I^{2} C} = ω^{2} x \end{array}$
Which also shows that rod is performing SHM