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Punchlinegirl
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A metal ball having net charge 8.9e-6 C is thrown out of a window horizontally at a speed 47 m/s. The window is at a height 89 m above the ground. A uniform horizontal magnetic field of magnitude 0.03 T is perpendicular to the plane of the ball's trajectory. Find the magnitude of the magnetic force acting on the ball just before it hits the ground. Answer in units of N.
I started off by using V_y^2= V_y_i^2 + 2a_y (y-y_i)
So the vertical component is -\sqrt 2gh j.
F_b= q v x B = q(v_i - 2 \sqrt 2gh j) x Bk = QvB(-j)-Q \sqrt 2gh B i
F_b = 8.9e-6 (47)(.03) j + (8.9e-6) \sqrt 2(9.8)(89)(.03)
F_b = 1.25e-5 j + 6.44e-5 i
Then I found the magnitude of F_b by squaring both terms and taking the square root to get 6.56e-5 N, but this wasn't right.. can someone help me? Thanks.
I started off by using V_y^2= V_y_i^2 + 2a_y (y-y_i)
So the vertical component is -\sqrt 2gh j.
F_b= q v x B = q(v_i - 2 \sqrt 2gh j) x Bk = QvB(-j)-Q \sqrt 2gh B i
F_b = 8.9e-6 (47)(.03) j + (8.9e-6) \sqrt 2(9.8)(89)(.03)
F_b = 1.25e-5 j + 6.44e-5 i
Then I found the magnitude of F_b by squaring both terms and taking the square root to get 6.56e-5 N, but this wasn't right.. can someone help me? Thanks.
