Question

# A -5.00 μC charge is moving at a constant speed of 6.90×105 m/s in the +x−direction...

A -5.00 μC charge is moving at a constant speed of 6.90×105 m/s in the +x−direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points.

Part A

x=0.500m,y=0, z=0

Bx,By,Bz =

nothing

T

Part B

x=0, y=0.500m, z=0

Bx,By,Bz =

nothing

T

Part C

x=0.500m, y=0.500m, z=0

Bx,By,Bz =

nothing

T

Part D

x=0, y=0, z=0.500m

Bx,By,Bz =

nothing

T

i,j and k are unit vector towards x, y, and x-direction.

Magnetic filed due to moving charge,

v = (6.9E5î + 0ĵ + 0k̂) m/s
r = (0.5î + 0ĵ + 0k̂) m
r² = (0.5² + 0² + 0²) = 0.25
|r| = 0.5

we know that cross product of two vectors in the same direction is 0. So,

so, B=0 and Bx=0, By=0 and Bz=0

part(B)

r = (0î + 0.5ĵ + 0k̂) m
r² = (0 + 0.5² + 0²) = 0.25
|r| = 0.5

So, magnetic field

Bx=0, By=0 and

part(C)

r = (0.5î + 0.5ĵ + 0k̂) m
r² = ( 0.5²+ 0.5² + 0²) = 0.5
|r| =

so,

So, it has only component in z direction, Bx=0, By=0 and

part(d)

r = (0î + 0ĵ + 0.5k̂) m
r² = ( 0²+ 0² + 0.5²) = 0.25

|r| =0.5

So, Bx=0 and Bz=0 ,