An electron (charge -e) is traveling horizontally east with speed v m/s. It is a

Question

An electron (charge -e) is traveling horizontally east with speed v m/s. It is about to hit a horizontal wire carrying current i amps North. what is the magnetic vector field (give direction and magnitude) acting on the electron when it is d meters from the wire? What is the magnitude and direction of the force?

I know the magnetic field is B = I/2r and the Force is = iLB

What I can't seem to understand is the right hand rule, there are a ton of variations so Im a little confused. Help is much appreciated. Thanks!

Explanation / Answer

   As stated, magnetic field   B   =   0 I / 2 r    Substituting   r   =   d    Magnetic field at the position of wire   B   =   0 I / 2 d    Direction of magnetic field can be calculated using right hand thumb rule:    Grasp the wires in right hand such that the thumb points in the direction of flow of current, then the fingers will show the direction of magnetic field.    Using this rule, the direction of magnetic field will be vertically outwards (perpendicular to the plane of paper).    Further force on a charged particle is      F   =   B q v sin    Since the direction of motion (due east) and magnetic field (upwards) are perpendicular, = 900.    =>   F   =   B q v                =   (0 I / 2 d) * e * v                =   0 I e v / 2 d               N    For the direction, stretch the right hand palm such that the fingers point in direction of field( i.e. upwards) and the thumb is in direction of current ( due west as the current flows in direction opposite to the direction of motion of electron) then the normal to the palm will point in the direction of magnetic force.    Hence the direction of magnetic force will be due north.

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