How Does Magnetic Force Affect a Charged Metal Ball in Motion?

[Punchlinegirl]

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.

 

[Punchlinegirl]

Can someone tell me what I'm doing wrong?

 

[kyle8921]

Your calculations seem to be correct, but there may be a small error in your calculation of the vertical component of the velocity. It should be -\sqrt{2gh} instead of -\sqrt{2gh}j. Also, make sure to convert all units to SI units (m, s, kg, etc.) before plugging them into the equation.

Alternatively, you can also use the formula F_b = qvBsin(theta), where theta is the angle between the velocity and the magnetic field. In this case, theta would be 90 degrees since the ball is moving horizontally and the magnetic field is perpendicular to it. So the magnitude of F_b would be qvB, which is (8.9e-6)(47)(0.03) = 1.25e-5 N.

In both cases, the magnitude of the magnetic force is 1.25e-5 N, which is the same as your calculated value. Make sure to double check your units and any small errors in calculations.