0-410 points SerPOP5 22 P033 M One long wire carries current 34.0 A to the left

Question

0-410 points SerPOP5 22 P033 M One long wire carries current 34.0 A to the left along the x axis. A second long wire carries current 76.0 A to the right along the line (y-0.280 m,z0) (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of-2.00 C is moving with a velocity of iSO.1m/s along the line (y-0.100 m, Z-0). Calculate the vector magnetic force acting on the relativistic effects.) (c) A uniform electric fieid is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field. N/C Need Help? Meste

Explanation / Answer

as the currents are in opposite directions

the magnetic fields directions are in opposite outside the line joinig the wires

34 A < 76 A

total magnetic field is zero below the wire caaying 32 A

magnetic field due to wire B = uo*I/(2*pi*r)

I1 = 34 A

B1 = uo*I1/(2*pi*r1) (out ward)

for I2 = 76 A

B2 = uo*I2/(2*pi*r2) ( inward)

r2 = r1 + y

Bnet = 0

B1 - B2 = 0

B1 = B2

I1/r1 = I2/(r1+y)

34/r1 = 76/(r1+0.28)

r1 = 0.23

at location y = -0.23 m <<<------ANSWER

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(b)

net magnetic field B = uo*I1/(2*pi*r1) + uo*I2/(2*pi*r2) into the page

r1 = 0.1 m

r2 = 0.28- 0.1 = 0.18 m

B = 4*pi*10^-7*34/(2*pi*0.1) + 4*pi*10^-7*76/(2*pi*0.18) -k

B = - 1.524*10^-4 T k

magnetic force Fb = q*( v x B)

Fb = -2*10^-6*(150*10^6 i x - 1.524*10^-4 k )

Fb = -2*10^-6*150*10^6*1.524*10^-4 j

Fb = -0.04572 j N

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Fnet = 0

Fb + Fe = 0

Fe = E*q

E = Fb/q

E = -0.04572/(2*10^-6) j

E = -22860 j N/C

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