A long, straight wire lies along the y axis and carries a current I-8.75 A in th

Question

A long, straight wire lies along the y axis and carries a current I-8.75 A in the -y direction (see the following tigure). In addition to the magnetic tield due to the current in the wire, a uniform magnetic field B with magnitude 1.15x10 6 T is in the +x direction. What is the total field (magnitude and direction) at the following points in the xz plane. (Give the smallest such angles.) Bo direction ofrom an axis parallel to the tx-direction in the xz plane (b) x 5.00 m, z -0 magnitude from an axis parallel to the x direction in the xz plane ()x 0, z - 1.25 Tm magnitude o from an axis parallel to the x-direction in the xz plane

Explanation / Answer

a) at x = 0, z = 5 m

magnetic field due to the wire,

B = mue*I/(2*pi*d)

= 4*pi*10^-7*8.75/(2*pi*5)

= 3.5*10^-7 T

= 0.35*10^-6 T (towards -x axis)

Bnet = Bo - B

= 1.45*10^-6 - 0.35*10^-6

= 1.1*10^-6 T

direction : 0 degrees

b)
at x = 5m , z = 0

magnetic field due to the wire,

B = mue*I/(2*pi*d)

= 4*pi*10^-7*8.75/(2*pi*5)

= 3.5*10^-7 T

= 0.35*10^-6 T (towards -x axis)

Bnet = sqrt(Bo^2 + B^2)

= sqrt(1.45^2 + 0.35^2)*10^-6

= 1.49*10^-6 T

direction : theta = tan^-1(B/Bo)

= tan^-1(0.35/1.45)

= 13.6 degrees

c)

at x = 0, z = -1.25 m

magnetic field due to the wire,

B = mue*I/(2*pi*d)

= 4*pi*10^-7*8.75/(2*pi*1.25)

= 1.4*10^-6 T   (towards -x axis)

Bnet = Bo + B

= 1.45*10^-6 - 1.4*10^-6

= 2.85*10^-6 T

direction : 0 degrees

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