A long straight wire is located at the origin and carries a current I1 = 3A in t

Question

A long straight wire is located at the origin and carries a current I1 = 3A in the direction shown. A 2nd long straight wire at x=3m, y-0m carries a current I2 = 4A in the direction shown. The field point has coordinates x=0, y=4. Find the magnitude and direction of the magnetic field at the field point P due to I_1. Express in i, j, k format. Find the magnitude and direction of the magnetic field at the field point due to I_2. Express in i, j, k format. Find the direction of the force on an electron placed at "P" due to WIRE 1 if the velocity is in the x direction Find the magnitude of the total magnetic field at P due to wire 1 and wire 2 A 8.0 mu C (magnitude) charge is traveling with a speed of 4.0 middot 10^5 m/s when it enters a magnetic field of strength 1.5 T. For each situation below do the following use traditional Cartesian coordinate system where the X-Y plane is the plane of the paper and "+i" is the right and "+j" is up Express V, B in i j, k format Determine the magnetic force on the charge expressed in i, j, k format Draw a picture showing the directions of V, B and F. Can use 2 d picture and "O" and "X" Use "O" and "X" as symbols for out of and into the plane of the paper respectively Negative Charge V into plane of paper B upwards Positive Charge V is in the Z-Y plane at an angle of 30 degrees off the positive "y-axis" in the 1st quadrant B into plane of paper Negative Charge V to the left B in the Z-Y plane at 60 degrees off the negative Z axis in the 2nd quadrant

Explanation / Answer

5)

A)

magnetic field at P due to I1:

B1 = -u*I1/(2*pi*d) i

= -1.26*10^-6*3/(2*pi*3) i

= (-2*10^-7T) i

B)

Similarly due to I2,

B2 = u*I2/(2*pi*r)

= 1.26*10^-6*4/(2*pi*5)

= (1.6*10^-7 T)*(4/5 i + 3/5 j )

= (6.42 i + 4.81 j )*10^-8 T

C)

As the direction of velocity and magnetic field are anti-parallel , the magnitude of magnetic force = 0

D)

Net magnetic field at P

Bnet = B1 + B2

=( (-20+6.42) i + 4.81 j)*10^-8

= (13.58 i + 4.81 j)*10^-8 T

6)

Magnetic force is given by:

F = q*( v x B)

where x denotes the vector product of v and B

A) for v into plane of paper : direction is -k

B is upwards .. direction: j

q is negative

So, direction of F = -i (towards Left)

B)

Here F = q*(cos(30) j + sin(30) k) x (-k)

= q*v*B*cos(30) i

So, direction is towards right

C)

Similarly here : F will be directed in y- Z axis

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