A copper bar of mass m rest at right angles to two parallel horizontal l distance apart . The rails are connected by a resistor R at one end and kept open at the other ends . Ther is a uniform upward magnetic fields of induction B.The bar is pulled away from the closed end by a constant force F. Calculate the terminal velocity of the bar when `mu` is the coefficient of frction between the rails and the bar .

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Correct Answer - `((F - mu mg)R)/(B^(2)l^(2))`

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