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Question

I am stuck on this problem and would be really greatful with any help on it.

A straight wire of mass 10.3 grams and length 5 cm is suspended from two identical springs that form a closed circuit (see figure below). The springs stretch a distance of 0.48 cm under the weight of the wire. The circuit has a total resistance of 10 Ω. When a magnetic field directed out of the page (indicated by the dots in the figure) is turned on, the springs are observed to stretch an additional 0.30 cm. What is the strength of the magnetic field? (The upper portion of the circuit is fixed.)

http://www.webassign.net/sercp/p19-66.gif

Answer 1

Pertinent formula to use,

F = BIL sin θ.

In this formula, F is the (magnetic) force exerted on a wire, lenght L, carrying current I. Current is perpendicular to field, so θ = 90°, sin 90° = 1; forget all about sin θ.

1. Find current. I = 24 V / 10 Ω = 2.4 A.

2. Determine magnetic force. This is the most intricate step. Find first spring constant k. For convenience, assume there's a single equivalent spring holding the wire. 10. 3 g = 10.3 E-3 kg; this mass weighs mg = 10.3 E-3 × 9.8 = 0.101 N. Spring elongation is 4.8 mm = 4.8 E-3 m. By Hooke's law F= kx, so k = F/x = 0.101/4.8 E-3 = 21.03 N/m. Also, from Hooke's law, dF = k dx. Thus additional force on spring, dF = k × additional elongation dx; dF = 21.03 × 3 E-3 = 0.06309 N. This additional force is that due to magnetic field on current-carrying wire.

3. Solve for B in the formula above. B = F/IL. Bear in mind that here F stands for magnetic force, only. Thus, B = 0.06309/(2.4 × 5 E-2) = 0.525 73 T.

Answer 2

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Answer 3

The magnetic field is 0.53 Tesla. I would be happy to go over the solution with you if you want to email me at the following address: jdlawlis at lycos.com. You will want to replace "at" with "@." (I am avoiding the spam bots who mine for email addresses).