A conducting rod of mass m and length l is suspended horizontally with the help of two identical non- conducting springs each of stiffness k in presence of uniform horizontal magnetic field of induction B. The ends of the rod are connected in a circuit consisting of an ideal battery of electromotive force E , a capacitance C and a switch as shown in the figure. Intially the switch was at position 1 for a long time and the switch is thrown to position 2 at $t = 0$ . ( neglect resisatance and inductance in the circuit and Consider the wires connecting the rod extremely flexible and denote acceleration due to gravity by g.)
Choose the correct alternatives

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

Conducting rod performs simple harmonic motion with angular frequency $ω = \sqrt{\frac{k}{m}}$ .

Conducting rod performs simple harmonic motion with angular frequency $ω = \sqrt{\frac{2k}{m}}$

Amplitude of simple harmonic motion of the conducting rod is $\frac{B l CE}{\sqrt{2km}}$ .

The conducting rod performs oscillatory motion but not simple harmonic motion.

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

SOL: Since resistance of rod is negligible capacitor instantaneously discharge through the
rod leading to impulse at equilibrium position given ${mv}_{max} = \int Bildt$
${mv}_{max} {=BlcE i.e v}_{max} = \frac{Blce}{m}$
But $v_{max} =Aw$
$\frac{blce}{m} =A \sqrt{\frac{2k}{m}}; A= \frac{Blce}{\sqrt{2km}}$