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Chapter 29, Problem 21P

(a)

To determine

The distance at which the magnitude of magnetic field is 0.1μT .

(a)

Expert Solution
Check Mark

Answer to Problem 21P

The distance at which the magnitude of magnetic field is 0.1μT is 4.00m .

Explanation of Solution

Given info: The magnitude of magnetic field is 1.00μT , the distance from the wire is 40.0cm and the current in each wire is 2.00A .

The formula to calculate the magnetic field is,

B=μI2πr (1)

Here,

I is the electric current in wire .

r is the distance from the wire.

μ is the permeability of free space.

Deduce the formula to calculate the distance from equation (1).

r=μI2πB (2)

Substitute 0.1μT for B , 2.00A for I and 4π×107T.m/A for μ in the above equation to find the value of r .

r=(4π×107T.m/A)2.00A2π(0.1μT×106T1μT)=4.00m

Thus, the distance at which the magnitude of magnetic field is 0.1μT is 4.00m .

Conclusion:

Therefore, the distance at which the magnitude of magnetic field is 0.1μT is 4.00m .

(b)

To determine

The magnetic field 40.0cm away from the middle of the straight cord.

(b)

Expert Solution
Check Mark

Answer to Problem 21P

The magnetic field 40.0cm away from the middle of the straight cord is 7.50nT .

Explanation of Solution

Given info: The magnitude of magnetic field is 1.00μT , the distance from the wire is 40.0cm and the current in each wire is 2.00A .

Thus, the direction of the magnetic force per unit length on a wire located 0.20cm from the center of the bundle is towards the center as the force is attractive in nature.

The formula to calculate the magnetic field is,

B=μI2πr (1)

Here,

I is the electric current in wire .

r is the distance from the wire.

μ is the permeability of free space.

The formula to calculate the net magnetic field is,

Bnet=B2B1 

Bnet=μI2πr2μI2πr1 (1)

The distance is measured from the center. The value of r1 is calculate as,

r1=40cm+12(3mm×101cm1mm)=40.15cm

The value of r2 is calculate as,

r2=40cm12(3mm×101cm1mm)=39.85cm

Substitute 4π×107T.m/A for μ , 2.00A for I , 40.15cm for r1 and 39.85cm for r2 in equation (1) to find the value of Bnet .

Bnet=(4π×107T.m/A)(2.00A)2π(39.85cm×102m1cm)(4π×107T.m/A)(2.00A)2π(40.15cm×102m1cm)=7.50×109T×109nT1T=7.50nT

Thus, the magnetic field 40.0cm away from the middle of the straight cord is 7.50nT .

Conclusion:

Therefore, the magnetic field 40.0cm away from the middle of the straight cord is 7.50nT .

(c)

To determine

The distance at which the magnetic field is one tenth of 7.50nT .

(c)

Expert Solution
Check Mark

Answer to Problem 21P

The distance at which the magnetic field is one tenth of 7.50nT is 1.26m .

Explanation of Solution

Given info: The magnitude of magnetic field is 1.00μT , the distance from the wire is 40.0cm and the current in each wire is 2.00A .

The formula to calculate the net magnetic field is,

Bnet=B2B1=μI2πr2μI2πr1  . (1)

The new value of Bnet is,

Bnet=Bnet10

Substitute 7.50nT for Bnet in the above equation to find the value of Bnet .

Bnet=7.50nT10=0.75nT

Consider a distance x at which the magnetic field is one tenth. The value of r1 is calculated as,

r1=xcm+12(3mm×101cm1mm)=(x+0.15)cm

The value of r2 is calculated as,

r2=xcm-12(3mm×101cm1mm)=(x0.15)cm

Substitute 4π×107T.m/A for μ , 2.00A for I , 0.75nT for Bnet , (x+0.15)cm for r1 and (x0.15)cm for r2 in  equation (1) to find the value of x .

(0.75nT)=(4π×107T.m/A)(2.00A)2π((x0.15)cm)(4π×107T.m/A)(2.00A)2π((x+0.15)cm)(0.75nT×109T1nT)=4×107(1(x0.15)1(x+0.15))0.00187=0.3x20.152x20.0225=160

Further solve for x .

x=(12.6cm×101m1cm)x=1.26m

Thus, the distance at which the magnetic field is one tenth of 7.50nT is 1.26m .

Conclusion:

Therefore, The distance at which the magnetic field is one tenth of 7.50nT is 1.26m .

(d)

To determine

The magnetic field the cable create at points outside the cable.

(d)

Expert Solution
Check Mark

Answer to Problem 21P

The magnetic field that the cable creates at points outside the cable is 0 .

Explanation of Solution

The magnitude of magnetic field is 1.00μT , the distance from the wire is 40.0cm and the current in each wire is 2.00A .

The formula to calculate the net magnetic field is,

Bnet=B2B1=μI2πr2μI2πr1 

The center wire in a coaxial cable carries a current is 2.00A and the current in the sheath around is 2.00A in the opposite direction. The radius of the center wire is almost equal to that of the sheath around it as the radius of the sheath is negligible. So the value of current and distance is same due to which the value of Bnet is zero.

Bnet=μI2πrμI2πr=0 

Thus, the magnetic field that the cable creates at points outside the cable is 0 .

Conclusion:

Therefore the magnetic field that the cable creates at points outside the cable is 0 .

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Chapter 29 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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