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.(adsbygoogle = window.adsbygoogle || []).push({});

I started off by using [tex] V_y^2= V_y_i^2 + 2a_y (y-y_i) [/tex]

So the vertical component is -[tex] \sqrt 2gh [/tex] j.

[tex] F_b= q v x B = q(v_i - 2 \sqrt 2gh j) x Bk = QvB(-j)-Q \sqrt 2gh B i [/tex]

[tex] F_b = 8.9e-6 (47)(.03) j + (8.9e-6) \sqrt 2(9.8)(89)(.03) [/tex]

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Magnetic force problem

**Physics Forums | Science Articles, Homework Help, Discussion**