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A straight wire of length L, mass m, carrying current I is suspended from a fixed point O. A second infinitely long wire carrying the same current I runs parallel to the first wire, located at a distance L below the lower end of the suspended wire. Both currents flow in the same direction. Given I = 2 A, L = 1 m, m = 0.1 kg, log 2 = 0.7. Find the angular acceleration of the suspended wire immediately after it is released.
- 48 * 10⁻⁷ rad/s²
- 96 * 10⁻⁷ rad/s²
- 24 * 10⁻⁶ rad/s²
- 48 * 10⁶ rad/s²
Correct answer: 96 * 10⁻⁷ rad/s²
Solution
An element at distance x from O is at distance (2L - x) from the infinite wire. The magnetic attraction per unit length is mu0 * I² / (2 * pi * (2L - x)). Integrating x * dF from 0 to L gives the net torque. The integral of x / (2L - x) dx evaluates to L(2 ln 2 - 1). With given numbers, torque = 3.2 * 10⁻⁷ N m. Dividing by moment of inertia (mL²/3 = 1/30) gives angular acceleration = 96 * 10⁻⁷ rad/s².
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