O 3/8/2017 11:55 PM A 66.7/100 Gradebook Print Calculator Periodic Table Questio

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O 3/8/2017 11:55 PM A 66.7/100 Gradebook Print Calculator Periodic Table Question 3 of 7 ncorrect ncorrect Map A 2a Sapling Learning macmillan lear wo long parallel wires carry currents of 2.65 A and 8.85 A. The magnitude of the force per unit length acting on each wire is 3.13 x 10 N/m. Find the separation distance of the wires expressed in millimeters Number 0.149856 Imam There is a hint available! View the hint by clicking on the bottom divider bar. Click on the divider bar again to hide the hint. Close Previous Check Answer Next Exit 8 Hint

Explanation / Answer

Q-3

a) Given Data

Currents I1 =2.65 A and I 2 = 8.85 A

force per unit lengh acting , F/l = 3.13*10^-5 N/m.

Solution :-

Force per unit length, F/l = 0*I1*I2 / (2d)
3.13*10^-5 = [ (4**10^-7)*(2.65*8.85) ] / [ (2**d) ]
3.13*10^-5= [ (2*10^-7)*(2.65*8.85) / (d) ]

3.13*10^-5= [ (4.69*10^-6) / (d)

Therefore separation distance, d = 0.1498 m = 149.85 mm

Q-7)

Given Data

Two "hemispheres" each about 70 mm wide.

one hemisphere = circular loop, 65.0 mm in diameter, => Radius, R = 65/2 = 32.5 mm = 0.0325 m

Magnetic Field , B = 4.55 fT = 4.55*10^-15 T

distance , x = 2.30 cm = 0.023 m

Solution : -

Magnetic Field , B = [ 0 / (4) ] * [ (2*I*R^2) / (R^2 + x^2)^1.5 ]

4.55*10^-15 = [ (1*10^-7) ] * [ (2*I*0.0325^2) / (0.0325^2 + 0.023^2)^1.5 ]

4.55*10^-15 = [ 1.05*10^-5 ] * I

=> I = 4.326*10^-10 A

At the center of hemisphere , x = 0

B = [ 0 / (4) ] * [ (2*I*R^2) / (R^2 + 0^2)^1.5 ]

B = [ (1*10^-7) ] * [ (2*4.326*10^-10*0.0325^2) / (0.0325^2 + 0.^2)^1.5 ]

B = 8.36*10^-15 T = 8.36 fT

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