{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

motion_in_E_and_B_fields

# motion_in_E_and_B_fields - opposite electric charge but the...

This preview shows pages 1–2. Sign up to view the full content.

Motion in E and B fields Given: C q 5 . 2 v Find: ? F s m v / 49 0 ? x E ? T B 3 . 3 y E ? z E ? x E ? y E ? z E A charged particle ( q = +2.5 μ C) moves at speed v 0 = 49 m/s in the + x -direction. At x = 0 it enters a region where a constant magnetic field B = 3.3 T is directed in the + z direction as shown in the diagram below. (The y -axis points into the screen.) (a) What is the magnitude of the magnetic force on the particle? Now a uniform electric field is applied so that the net force on this particle is zero. At the moment the particle first enters the region of the magnetic field, what direction is the force due to the magnetic field on the particle? Direction of force due to magnetic field? (b) What are the components of the electric field? (c) If the incoming particle were replaced by one with the

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: opposite electric charge but the same initial velocity, what electric field would be required to ensure a straight line motion? (Magnetic field is unchanged.) Solutions a.) To find F we can use the following formula: qvB B v q F using the right hand rule you get the direction to be -y qvB F b.) To find , & we can use the following formula: x E y E z E qE F q F E vB q qvB E Since the force is only in the y direction we get: & z x E E vB E y c.) To find , & we can use the following formula: x E y E z E vB q qvB E Since the force is only in the y direction we get: & z x E E vB E y...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

motion_in_E_and_B_fields - opposite electric charge but the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online