The flux graph through the loop.

Question

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Answer 1

Question

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

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Answer 2

Question

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

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Answer 3

Question

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

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Answer 4

Question

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

Answer 6

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

Answer 7

Chapter 28, Problem 39P

(a)

To determine

The flux graph through the loop.

Answer 8

(a)

Expert Solution

Answer to Problem 39P

The flux graph through the loop as a function of time is shown in figure 1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The length of the rectangular loop, $Lside of theloop=10 cm$

The widthof the rectangular loop, $r=5 cm$

The stance of the rectangular loop, $R=2.5 Ω$

The speed of the rectangular loop, $v=2.4 cm/s$

The magnetic field of the rectangular loop, $B=1.7 T$

Formula used:

The formula for flux through the loop as a function of time is given by,

$ϕm=mt+b$

The expression for the time required to enter into uniform magnetic field is given by,

$t=Lside of theloopv$

The expression of flux in terms of the length is given by,

$ϕm=BrLside of the loop$

Calculation:

The time required to enter into uniform magnetic field is calculated as,

$t=L side of theloopv=( 10 cm( 10 −2 m 1 cm ))2.4 cms( 10 −2 m 1 cm )=0.10 m2.4× 10 −2 m/s=4.17 s$

The flux $(4<t<17 s)$ through the loop can be calculated as,

$ϕm=BrLside of the loop=(1.7 T( Wb/ m 2 1 T ))×(5 cm( 10 −2 m 1 cm ))×(10 cm( 10 −2 m 1 cm ))=(1.7 Wb m 2 )(0.05 m)(0.10 m)=(8.5× 10 −3Wb)$

For, $t=8.33 s$ , the flux is calculated as,

$8.50×10−3 Wb=m×8.33 s+b$ ..... (1)

For, $t=12.5 s$ , the flux is calculated as,

$ϕm=mt+b0 Wb=m×12.5 s+b$ ...... (2)

From equation (1) and (2)

$m=2.04×10−3 Wbb=22.5×10−3$

Now,

$ϕm=mt+b=(2.04× 10 −3 Wb)t+22.5×10−3$

The induced current through the loop as a function of time is shown in figure 1.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 1

Figure 1

Conclusion:

The graph of induced emf through the loop as a function of time is shown in the figure 1.

Answer 13

(b)

To determine

The graph of induced emf and current through the loop.

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

Answer 14

(b)

To determine

The graph of induced emf and current through the loop.

Answer 15

(b)

Expert Solution

Answer to Problem 39P

The graph of induced emf and current through the loop is shown in the figure 2 and 3.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The formula for induced emfthrough the loop is given by,

$ε=−dϕmdt$

Calculation:

The induced emf through the loop when $t>7 s$ is calculated as,

$ε=−dϕmdt=−d( 2.04 mWb s )dt=−d( 2.04 mWb s ( 1 V⋅s 1 Wb ))dt=−2.04 mV$

Theinduced emfthrough the loop when $t>12.5 s$ is calculated as,

$ε=−tdϕmdt−ddt(( 2.04 mWb s ( 1 V⋅s 1 Wb ))+( 25.5 × 10 −3 ))−ddt(( 2.04 mV)+( 25.5 × 10 −3 ))=−2.04 mV$

The current through the loop when $t>12.5 s$ can be defined as,

$ε=−2.04 mV$

The graph for the emf is shown in figure 2.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 2

Figure 2

The expression for the value of current is given by,

$I=εR$

The graph for the current is shown in figure 3.

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 28, Problem 39P , additional homework tip 3

Figure 3

Conclusion:

The graph of induced emfand current through the loop is shown in the figure 2 and 3.

Answer 20

Ch. 28 - Prob. 1P Ch. 28 - Prob. 2P Ch. 28 - Prob. 3P Ch. 28 - Prob. 4P Ch. 28 - Prob. 5P Ch. 28 - Prob. 6P Ch. 28 - Prob. 7P Ch. 28 - Prob. 8P Ch. 28 - Prob. 9P Ch. 28 - Prob. 10P

The flux graph through the loop.

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The flux graph through the loop.

Answer to Problem 39P

Explanation of Solution

The graph of induced emf and current through the loop.

Answer to Problem 39P

Explanation of Solution

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