The magnitude and direction of induced emf, induced current, and force on the loop.

Question

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Answer 1

Question

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

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Answer 2

Question

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

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Answer 3

Question

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

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Answer 4

Question

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

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Answer 5

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Answer 6

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Answer 7

Chapter 20, Problem 89P

(a)

To determine

The magnitude and direction of induced emf, induced current, and force on the loop.

Answer 8

(a)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The loop is at rest and magnetic field is constant, and average velocity of electron is also zero

Write the expression for magnetic force

$FB=q(v×B)$ (I)

Here, $FB$ is the magnetic force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength

Since velocity of electrons is zero, according to equation (I) the magnetic force will be zero.

Write the expression for induced emf

$ε=(v→×B→)∥L$ (II)

Here, $L$ is the length of loop, $ε$ is the induced emf

From equation (II) when velocity of electrons become zero, the emf is also become zero.

The expression for induced current

$I=εR$ (III)

Here, $R$ is the resistance of loop

Since emf is zero, induced current is also zero according to equation (I)

Conclusion:

Therefore, the magnitude of induced emf is $ε=0_$ , induced current is $I=0_$ , and magnetic force is $zero_$ .

Answer 13

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Answer 14

(b)

To determine

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

Answer 15

(b)

Expert Solution

Answer to Problem 89P

The magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

In this case electrons are moving through the loop and there is a magnetic field associated with the motion of electrons. According to right hand rule, when a magnetic field is in the direction shown by curled fingers, the direction of force is along the right thumb. Since negatively charged electrons moves to right, the direction of field is downward in the plane of paper.

Write the expression for magnetic field in a loop

$B→=μ0I2πr$ (IV)

Here, $I$ is the current in loop, $B$ is the magnetic field, $r$ is the radius of the loop

The expression for magnetic force is

$F→=−ev→×B→$ (V)

The force will develop an emf in loop in the left and right side, since left side is close to the wire, the magnetic field will be greater at right side, hence electron circulate in counterclockwise direction, and current flows in clockwise direction.

Write the expression for induced emf

$εnet=vL(BL−BR)$ (VI)

Here, $BL$ is the magnetic field at left, and $BR$ is the magnetic field at right

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (VI)

$εnet=μ0I2πvL(1RL−1RR)$ (VII)

The current in the loop is

$Iloop=εnetR$ (VIII)

The magnetic force acting on each segment of loop be $F→=IloopL→×B→$ , the top and bottom of loop experiences equal and opposite forces, and force on left acting to left and right to right, with force on left is greater.

The expression for magnitude of force is

$∑F=IloopL(BL−BR)$ (IX)

Substitute, $μ0I2πRL$ for $BL$ , and $μ0I2πRR$ for $BR$ in equation (IX)

$∑F=μ0ILIloop2π(1RL−1RR)$ (X)

Conclusion:

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , and $0.113 m$ for $RR$ in equation (VII)

$εnet=4π×10−7 T⋅m/A×6.8 A2×3.14(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=32 nV$

Substitute, $3.183×10−8 V$ for $εnet$ , and $79 Ω$ for $R$ in equation (VIII)

$Iloop=3.183×10−8 V79 Ω=400 pA$

Substitute, $4π×10−7 T⋅m/A$ for $μ0$ , $6.8 A$ for $I$ , $0.45 m/s$ for $v$ , $0.023 m$ for $L$ , $0.090 m$ for $RL$ , $4.03×10−10 A$ for $Iloop$ , and $0.113 m$ for $RR$ in equation (X)

$∑F=4π×10−7 T⋅m/A×6.8 A2×3.14(4.03×10−10 A)(0.45 m/s)(0.023 m)(10.090 m−10.113 m)=2.9×10−17 N$

The force is acting to left.

Therefore, the magnitude of induced emf is $εnet=32 nV_$ , induced current is $Iloop=400 pA_$ , and magnetic force is $2.9×10−17 N_$ .

Answer 20

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Answer 21

(c)

To determine

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

Answer 22

(c)

Expert Solution

Answer to Problem 89P

The electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for power dissipated in the loop

$PE=Iε$ (XI)

Here, $I$ is the induced current, $ε$ is the induced emf

Write the expression for power dissipated by external force

$PF=Fv$ (XII)

Here, $F$ is the magnetic force, $v$ is the velocity of electrons

Conclusion:

Substitute, $4.03×10−10 A$ for $I$ , and $3.183×10−8 V$ for $ε$ in equation (XI)

$PE=(4.03×10−10 A)(3.183×10−8 V)=1.3×10−17 W$

Substitute, $2.85×10−17 N$ for $F$ , and $0.45 m/s$ for $v$ in equation (XII)

$PF=(2.85×10−17 N)(0.45 m/s)=1.3×10−17 W$

Thus, $PE=PF$ .

Therefore, the electric power dissipated in the loop is $PE=1.3×10−17 W_$ , and it is equal to power dissipated by external force.

Answer 27

Ch. 20.1 - 20.1 If the rod in Fig. 20.1 were moving out of...Ch. 20.1 - Conceptual Practice Problem 20.1 Loop of Different...Ch. 20.2 - Prob. 20.2CP Ch. 20.2 - Prob. 20.2PP Ch. 20.3 - Prob. 20.3PP Ch. 20.3 - Practice Problem 20.4 Rotating Coil Generator In a...Ch. 20.4 - Figure 20.11 Circular loop in a magnetic field of...Ch. 20.4 - Prob. 20.5PP Ch. 20.4 - (a) In Fig. 20.21, just after the switch is...Ch. 20.6 - CHECKPOINT 20.6 The primary coil of a...

The magnitude and direction of induced emf, induced current, and force on the loop.

Videos

The magnitude and direction of induced emf, induced current, and force on the loop.

Answer to Problem 89P

Explanation of Solution

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .

Answer to Problem 89P

Explanation of Solution

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

Answer to Problem 89P

Explanation of Solution

Want to see more full solutions like this?

Chapter 20 Solutions

The magnitude and direction of induced emf, induced current, and force on the loop.

Videos

The magnitude and direction of induced emf, induced current, and force on the loop.

Answer to Problem 89P

Explanation of Solution

Repeat part (a) for a loop moving to right with constant speed 45 cm/s<m:math><m:mrow><m:mn>45</m:mn><m:mtext> </m:mtext><m:mtext>cm/s</m:mtext></m:mrow></m:math>.

Answer to Problem 89P

Explanation of Solution

The electric power dissipated in the loop, and show that it is equal to the rate of external force.

Answer to Problem 89P

Explanation of Solution

Want to see more full solutions like this?

Chapter 20 Solutions

Repeat part (a) for a loop moving to right with constant speed $45 cm/s$ .