The magnetic field.

Question

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Answer 1

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

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Answer 2

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

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Answer 3

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

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

Answer 4

Question

Definition Definition Angle at which a point rotates around a specific axis or center in a given direction. Angular displacement is a vector quantity and has both magnitude and direction. The angle built by an object from its rest point to endpoint created by rotational motion is known as angular displacement. Angular displacement is denoted by θ, and the S.I. unit of angular displacement is radian or rad.

Chapter 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

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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 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

Answer 6

Chapter 26, Problem 23P

To determine

The magnetic field.

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10

Hence the magnetic field is given as

B→=ai^+bj^+ck^B→=(10i^+10j^−15k^) T

Conclusion:

The magnetic field comes out to be (10i^+10j^−15k^) T .

Answer 7

Chapter 26, Problem 23P

To determine

The magnetic field.

Answer 8

Expert Solution & Answer

Answer to Problem 23P

(10i^+10j^−15k^) T

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical =θ

Magnitude of current in the wire =I=2 A

Length of the wire before rotating =li→=(0.10 m) i^ + (0 m) j^ + (0 m) k^

Length of the wire after rotating =lf→=(0 m) i^ + (0.10 m) j^ + (0 m) k^

Magnetic field =B→=ai^+bj^+ck^

Magnetic force before rotating =Fi→=(0 N)i^+(3 N)i^+(2 N)k^

Magnetic force after rotating =Ff→=(0 N)i^−(3 N)j^−(2 N)k^

Formula Used:

Magnetic force on a current carrying wire is given as

F→=I(l→×B→)

Calculation:

Magnetic force before rotating is given as

Fi→=I(li→×B→)(0)i^+(3)j^+(2)k^=(2)(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(0)i^+(1.5)j^+(1)k^=(((0.10 ) i^ + (0 ) j^ + (0 ) k^)×(ai^+bj^+ck^))(1.5)j^+(1)k^=(0.10 )bk^−(0.10 )cj^ comparing both side (0.10 )b=1 and −(0.10 )c=1.5b=10 and c = -15

Magnetic force after rotating is given as

Ff→=I(lf→×B→)(0)i^−(3)j^−(2)k^=(2)[((0) i^ + (0.10) j^ + (0) k^)×(ai^+bj^+ck^)](0)i^−(1.5)j^−(1)k^=[(0) i^ + (0.10) j^ + (0) k^]×(ai^+10j^−15k^)−(1.5)j^−(1)k^=(−(0.10)ak^)−(1.5)i^ comparing both side −(0.10)a=−1 a =10