a sort of projectile launcher is shown in fig 24.65. A large current moves in a

Question

a sort of projectile launcher is shown in fig 24.65. A large current moves in a closed loop composed of fixed rails, a power supply, and a very light, almost frictionless bar touching the rails. a magnetic field is perpendicular to the plane of the circuit. if the bar has a length L = 22cm, a mass of 1.5g, and is placed in a field of 1.7 T, what constant flow is needed to accelerate the bar from rest to 28 m/s in a distance of 1.0m? In what direction must the magnetic field point?

For some reason i cant attach the figure for this problem. If you can imagine a rectangle, the left short end contains the power supply while the right short end is the light bar of lenth L. The current is moving towards the right on the top of the rectangle whereas it is moving back to the left on the bottom, and downwards along the bar.

Explanation / Answer

Given that,    The length of the bar L = 22 cm                                         = (22 cm)(10-2 m/1 cm)                                         = 0.22 m    The mass of the bar m = 1.5 g                                        = (1.5 g)(10-3 kg/1 g)                                        = 0.0015 kg    The magnetic field B = 1.7 T    The velocity of the bar v = 28 m/s    The distance d = 1.0 m _______________________________________________ a)    The initial velocity v0 = 0 m/s    Using kinematic equation, we have                 v2 = v02+2as                  v2 = 0+2ad    Therefore, the acceleration is                   a = v2/2d                      = (28 m/s)2/(2)(1.0 m)                      = (28 m/s)2/(2)(1.0 m)                      = 392 m/s2 ________________________________________________    The magnetic force F = BIL     ...... (1)    From Newton's second law of motion,                                   F = ma    Therefore, the equation (1) becomes                               ma = BIL                                   I = ma/BL   ...... (2)    Using the equation (2), the current is                                  I = (0.0015 kg)(392 m/s2)/(1.7 T)(0.22 m)                                    = 1.57 A                                    = 1.6 A (Approximately) __________________________________________________ __________________________________________________    b)    The direction of the magnetic field:    Using right hand rule, the dirction of the magnetic field is along the downward.                                     I = ma/BL   ...... (2)    Using the equation (2), the current is                                  I = (0.0015 kg)(392 m/s2)/(1.7 T)(0.22 m)                                    = 1.57 A                                    = 1.6 A (Approximately) __________________________________________________ __________________________________________________    b)    The direction of the magnetic field:    Using right hand rule, the dirction of the magnetic field is along the downward.                      

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