Note: The positive z-axis is out of the page (towards you)...

Note: The positive z-axis is out of the page (towards you) and the negative z-axis is into the page (away from you). 21) (25 points) Three long wires carry currents I1, I2, and I3 parallel to the z-axis. Wire 1 is located at the origin and carries current I1 = 3.0 A in the positive z-direction. Wire I2 is located at x = 1.0 cm and carries 2.0 A of current in the negative z-direction. Wire I3 is located at x = 2.0 cm and carries 1.0 A of current in the negative z-direction. A negatively charged particle of magnitude q = 0.50 C is currently located at x = 3.0 cm and is moving in the positive x-direction at a speed of 1.0 × 10^6 m/s. What is the magnitude and direction of the net magnetic force on the particle? Note: The permeability of free space is ?0 = 4? × 10^-7 T·m/A

Transcript

00:01 Now the charge particle of charge q which is moving with a charge q is equal to 0 .5 column and this particle is moving with a velocity 10 -to -6 meter per second on the positive x -axis.

00:20 So we know the unit vector in the positive x -axis is plus i.

00:26 So its velocity can be shown directly with the direction is 10 -rest -to -6 -i meter per second.

00:38 Now we need to find out the total force acting on this particle.

00:42 We know when a charge particle is in a magnetic field, the force acts on the particle is equal to q into v -cross -b.

00:53 It is the vector product of velocity of the particle.

00:57 And the magnetic field in which the particle is.

01:01 So we can write it q, vb, sine theta.

01:08 So in order to find out the total magnetic field, q is known to us, b is known to us.

01:14 So we need to find out the total magnetic field at the location of the particle and we need to find out also the angle between the velocity of the particle and the magnetic field.

01:27 Means we need to find out the direction of the magnetic field because the direction of the particle is known so we can know the angle now total magnetic field at the position of the particle is due to a wire carrying current i1 we call this field as b1 and due to the wire carrying means current i 2 we call it as field b2 and due to the wire carrying means current i 3 we call it field b3 and we have a standard formula suppose this wire is carrying current i then at a distance r from it the value of the magnetic field b is equal to mu zero i upon 2 pi r where mu is 4 pi into 10 to 7 it is a constant and the direction of this magnetic field is given by the right -hand rule it is always perpendicular to the means plain carrying the wire either upward or down downward so use the formula to use the right -hand rule to find the direction so what will be b1 first we calculate what will be b1 b1 will be mu -0 i 1 upon 2 pi r1 and b2 will be mu 0 i 2 i 2 upon 2 pi r2 and b 3 is equal to mu 0 i 3 upon 2 pi r 3 now this is a simple magnitude now we have to find out its direction also so we will consider the the direction of the magnetic field now means from the right hand rule, if we keep our thumb in the direction of the flow of the current, we hold the wire keeping our thumb in the direction of the flow of the current, then the curved finger tells us the direction of the magnetic field...

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