標題:
physics
發問:
A sort of "projectile launcher" . 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. The bar has a L = 20 cm, a mass of 1.3 g, and is placed in... 顯示更多 A sort of "projectile launcher" . 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. The bar has a L = 20 cm, a mass of 1.3 g, and is placed in a field of 1.7 T. What constant current flow is needed to accelerate the bar from rest to 34 m/s in a distance of 1.0 m? 更新: piture:http://space.uwants.com/index.php/881818/action_viewspace_itemid_146973
最佳解答:
First of all, we have to find out the value of the acceleration that the bar needs: By v2 - u2 = 2as, put v = 34, u = 0 and s = 1: 1156 = 2a(1) a = 578 m s-2. Therefore, the megnetic force needed to be acted on the bar is given by: ma = (0.0013 × 578) = 0.7514 N Finally, using the equation of magnetic force: F = BIL where B = B-field, I = current and L = length of the segment carrying current, 0.7514 = 1.7 × I × 0.2 I = 2.21 A
physics
發問:
A sort of "projectile launcher" . 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. The bar has a L = 20 cm, a mass of 1.3 g, and is placed in... 顯示更多 A sort of "projectile launcher" . 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. The bar has a L = 20 cm, a mass of 1.3 g, and is placed in a field of 1.7 T. What constant current flow is needed to accelerate the bar from rest to 34 m/s in a distance of 1.0 m? 更新: piture:http://space.uwants.com/index.php/881818/action_viewspace_itemid_146973
最佳解答:
First of all, we have to find out the value of the acceleration that the bar needs: By v2 - u2 = 2as, put v = 34, u = 0 and s = 1: 1156 = 2a(1) a = 578 m s-2. Therefore, the megnetic force needed to be acted on the bar is given by: ma = (0.0013 × 578) = 0.7514 N Finally, using the equation of magnetic force: F = BIL where B = B-field, I = current and L = length of the segment carrying current, 0.7514 = 1.7 × I × 0.2 I = 2.21 A
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