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Period of a metal rod oscillating in a magnetic field
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[QUOTE="etotheipi, post: 6423689"] There's a long way and a short way to solve the problem (let's assume we can neglect any possible variation in current due to the change in flux linked by the configuration). You can either write the ##\tau = I\ddot{\theta}## equation, namely$$-mgl\sin{\theta} + BILl \cos{\theta} = ml^2 \ddot{\theta}$$And then let ##\theta = \theta_0 + \varepsilon##, where ##\varepsilon## is the (small) angular displacement from the equilibrium position. Then, you can show using the double angle formula that ##\sin{(\theta_0 + \varepsilon)} \approx \sin{\theta_0} + \varepsilon \cos{\theta_0}## as well as ##\cos{(\theta_0 + \varepsilon)} = \cos{\theta_0} - \varepsilon \sin{\theta_0}##, and if you plug this in, and use that ##\ddot{\theta} = \ddot{\varepsilon}##, you'd find you end up with something in the SHM form. However, there's a better way to solve the problem! Notice that a constant force of magnitude ##\sqrt{(mg)^2 + (BIL)^2}## acts on the bar, in a constant direction. If you tilt your head, then this is entirely equivalent to a simple pendulum swinging under the influence of an 'effective' gravitational acceleration ##\sqrt{g^2 + ((BIL)/m)^2}##! Does that suggest an easier solution? [/QUOTE]
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Period of a metal rod oscillating in a magnetic field
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