(7%) Problem 2: A rod of m = 2.6 kg rests on two parallel rails that are L = 0.3

Question

(7%) Problem 2: A rod of m = 2.6 kg rests on two parallel rails that are L = 0.34 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 4.3 A. A uniform magnetic field of magnitude B = 0.65 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d = 0.95 m. Ignore the friction on the rails ©theexpertta.com 4. Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times Grade Summary Deductions Potential 0% 100% sin Submissions Attempts remaining: Z (0% per attempt) detailed view tan() | | ( cosO cotan0 asin) acos0 atan0a acotans sinhO cosh0 tanh0 cotanh0 Degrees O Radians 0 END 0 DEL CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback

Explanation / Answer

As per the question, both parallel rails are apart with L = 0.34m and current in rod is i = 4.3 A then-

As we know that force on moving rod in magnetic feild is, F = BiL

F = 0.65* 4.3*0.34

F = 0.95N

then acceleration of rod will be -

F = ma

0.95 = 2.6(a)

a = 0.365 m/s2

Now after traveled s = 0.95 m, final velocity(V) will be-

V2 = U2 +2as

V2 = 0 + 2(0.365)(0.95) (U =0 because rod started from rest)

V2 = 0.694

V = 0.834 m/s

Thank you !!! Have fun

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.