EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100460300
Author: SERWAY
Publisher: YUZU
Question
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Chapter 31, Problem 31.13P

(a)

To determine

The magnetic flux through the loop due to the current.

(a)

Expert Solution
Check Mark

Answer to Problem 31.13P

The magnetic flux through the loop due to the current is μ0IL2πln(1+wh) .

Explanation of Solution

Given info: The length of the rectangle is 1.00m , the value of b is 10.0A/s , the value of h is 1.00cm and the value of width of the rectangle is 10.0cm .

The magnetic field is a function of the distance. So, the value of the magnetic field varies over the area of the rectangular loop.

Consider an infinitesimal section of the loop at a distance of x from the wire of thickness dx .

Write the expression for the area of the infinitesimal section of the loop.

dA=Ldx

Here,

L is the length of the rectangle.

dx is the thickness of the infinitesimal section.

Write the expression for the magnetic field at the infinitesimal section.

B=μ0I2πx

Here,

μ0 is the vacuum permeability.

x is the distance from the wire to the infinitesimal section.

I is the current flowing in the wire.

Write the expression for the magnetic flux through the infinitesimal section.

dϕ=BdA

Substitute μ0I2πx for B and Ldx for dA in above equation.

dϕ=(μ0I2πx)(Ldx)=μ0ILdx2πx

Write the expression for the magnetic flux through the entire loop.

ϕ=μ0IL2πhh+wdxx=μ0IL2π(lnx)hh+w=μ0IL2π(ln(h+w)lnh)=μ0IL2πln(1+wh)

Conclusion:

Therefore, the magnetic flux through the loop due to the current is μ0IL2πln(1+wh) .

(b)

To determine

The induced emf in the loop.

(b)

Expert Solution
Check Mark

Answer to Problem 31.13P

The induced emf in the loop is 4.80μV .

Explanation of Solution

Given info: The length of the rectangle is 1.00m , the value of b is 10.0A/s , the value of h is 1.00cm and the value of width of the rectangle is 10.0cm .

Write the expression for the induced emf in the loop.

ε=dϕdt

Here,

dt is the change in time.

From part (a), the magnetic flux through the loop due to the current is μ0IL2πln(1+wh) .

Substitute μ0IL2πln(1+wh) for ϕ in above equation.

ε=d(μ0IL2πln(1+wh))dt=μ0L2πln(1+wh)dIdt

Substitute a+bt for I in the above equation.

ε=μ0L2πln(1+wh)d(a+bt)dt=μ0L2πln(1+wh)(b)

Substitute 4π×107Tm/A for μ0 , 1.00m for L , 10.0cm for w , 1.00cm for h and 10.0A/s for b in above equation.

ε=(4π×107Tm/A)(1.00m)2πln(1+10.0cm1.00cm)(10.0A/s)4.79×106V×106μV1V4.80μV

Conclusion:

Therefore, the induced emf in the loop is 4.80μV .

(c)

To determine

The direction of the induced current in the rectangle.

(c)

Expert Solution
Check Mark

Answer to Problem 31.13P

The direction of the induced current in the rectangle is counterclockwise.

Explanation of Solution

Given info: The length of the rectangle is 1.00m , the value of b is 10.0A/s , the value of h is 1.00cm and the value of width of the rectangle is 10.0cm .

Write the expression for the current induced in the loop.

I=εR

Here,

R is the resistance.

From the above expression, the current is directly proportional to the emf induced in the loop. From part (b), the induced emf in the loop is 4.80μV . The value of the current is negative because the sign of the emf induction is negative which means the current flows in opposite direction from the direction which is shown in given figure. The negative sign indicates the counterclockwise direction by the right hand thumb rule. Thus, the current flows in the counterclockwise direction.

Conclusion:

Therefore, the direction of the induced current in the rectangle is counterclockwise.

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Chapter 31 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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