Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
11.3k views
in Physics by (80.9k points)

A rigid rod of mass m and length L is suspended from two identical springs of negligible mass as shown in the diagram above. The upper ends of the springs are fixed in place and the springs stretch a distance d under the weight of the suspended rod.

a. Determine the spring constant k of each spring in terms of the other given quantities and fundamental constants.

As shown above, the upper end of the springs are connected by a circuit branch containing a battery of emf ε and a switch S so that a complete circuit is formed with the metal rod and springs. The circuit has a total resistance R, represented by the resistor in the diagram. The rod is in a uniform magnetic field directed perpendicular to the page. The upper ends of the springs remain fixed in place and the switch S is closed. When the system comes to equilibrium, the rod has been lowered an additional distance Δd. 

b. With reference to the coordinate system shown above on the right, what is the direction of the magnetic field? 

c. Determine the magnitude of the magnetic field in terms of m, L, d, Δd, ε, R, and fundamental constants. 

d. When the switch is suddenly opened, the rod oscillates. For these oscillations, determine the following quantities in terms of d, Δd, and fundamental constants: 

i. The period 

ii. The maximum speed of the rod

1 Answer

+1 vote
by (52.5k points)
selected by
 
Best answer

(a) Using hookes law. 2Fsp = mg   2(kΔx) = mg   k = mg/2d

(b) From the battery, we can see that + current flows to the right through the rod. In order to move the rod down a distance Δd, the magnetic force must act down. Based on the RHR, the B field would have to act out of the page (+z).

(c) The extra spring stretch must be balanced by the magnetic force. Fsp(extra) = Fb

kΔd = BIL … (mg/2d Δd) = BIL Now substitute ε = IR for I and we get B  B = mgRΔd / εLd

(d)

(use 2k, for k since there are two springs)

ii) Set the equilibrium position (at Δx = d) as zero spring energy to use the turn horizontal trick. This is the maximum speed location and we now set the kinetic energy here to the spring energy and the Δd stretch position.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...