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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A 1.00-kg ball having net charge Q = 5.00 μC is thrown out of a window horizontally at a speed v = 20.0 m/s. The window is at a height h = 20.0 m above the ground. A uniform horizontal magnetic field of magnitude B = 0.010 0 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. Hint: Ignore magnetic forces in finding the ball’s final velocity.

To determine
The magnitude of the magnetic force acting on the ball just before it hits the ground.

Explanation

Given info: The mass of the ball is 1.00kg . The net charge in the ball is 5.00μC . The ball is thrown horizontally with speed 20.0ms-1 . The height of the window is 20.0m . The magnetic field perpendicular to the plane of ball trajectory has magnitude of 0.0100T .

Explanation:

The magnetic force on the ball is given by,

F=qvBsinθ

  • q is the charge on the ball
  • v is the velocity of the ball
  • B is the magnetic field
  • θ is the direction of the magnetic field and the velocity of the ball

Since the magnetic field is perpendicular to the plane of ball trajectory,

F=qvBsin90°=qvB       (1)

From equation of motion,

vy2vy02=2aΔy

  • vy is the velocity in the y direction
  • vy0 is the initial velocity in the y-direction
  • a is the acceleration in the y-direction
  • Δy is the displacement in the y-direction

Since the initial velocity in the y-direction is zero and the acceleration in the y-direction is acceleration due to gravity,

vy2=2gΔy

  • g is the free fall acceleration

The velocity in the y-direction will be,

vy=2gΔy       (2)

Since there is no force acting in the x-direction, the x-component of the velocity remains unchanged.

vx=vx0       (3)

The magnitude of the net velocity will be,

v=(vx)2+(vy)2       (4)

Combining (2), (3) and (4),

v=(vx0)2+(2gΔy)2       (5)

The magnetic force on the ball using (5) will be,

F=q((vx0)2+(2gΔy)2)B

Substituting 5

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