An election moves through a uniform electric field E → = (2.50 i ^ + 5.00 j ^ ) V/m and a uniform magnetic field B → = 0.400 k ^ T. Determine the acceleration of the electron when it has a velocity v → = 10.0 i ^ m/s.

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Answer 1

Textbook Question

Chapter 34, Problem 34.4P

An election moves through a uniform electric field E → = (2.50 i ^ + 5.00 j ^ ) V/m and a uniform magnetic field B → = 0.400 k ^ T. Determine the acceleration of the electron when it has a velocity v → = 10.0 i ^ m/s.

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

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Answer 2

Textbook Question

Chapter 34, Problem 34.4P

An election moves through a uniform electric field E → = (2.50 i ^ + 5.00 j ^ ) V/m and a uniform magnetic field B → = 0.400 k ^ T. Determine the acceleration of the electron when it has a velocity v → = 10.0 i ^ m/s.

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

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Answer 3

Textbook Question

Chapter 34, Problem 34.4P

An election moves through a uniform electric field E → = (2.50 i ^ + 5.00 j ^ ) V/m and a uniform magnetic field B → = 0.400 k ^ T. Determine the acceleration of the electron when it has a velocity v → = 10.0 i ^ m/s.

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

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Answer 4

Textbook Question

Chapter 34, Problem 34.4P

An election moves through a uniform electric field E → = (2.50 i ^ + 5.00 j ^ ) V/m and a uniform magnetic field B → = 0.400 k ^ T. Determine the acceleration of the electron when it has a velocity v → = 10.0 i ^ m/s.

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

Answer 8

Expert Solution & Answer

To determine

The acceleration of electron.

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .

The formula for force acting on an electron in an electric field is,

F→e=qE→ . (1)

Here,

q is the charge.

E→ is the electric field.

F→e is force due to electric field.

Substitute −1.6×10−19 C for q and (2.50i^+5.00j^) V/m for E→ in equation (1).

F→e=(−1.6×10−19 C)((2.50i^+5.00j^) V/m)=(−1.6×10−19 C)((2.50i^+5.00j^) ) N/C=(−4×10−19i^−8×10−19j^) N

The formula for force acting on an electron in an magnetic field is,

F→m=q(v→×B→) (2)

Here,

B→ is magnetic field.

v→ is velocity.

F→m is force due to magnetic field.

Substitute −1.6×10−19 C for q , 10.0i^ m/s for v→ and (0.400k^) T for B→ in equation (2).

F→m=(−1.6×10−19 C)((10.0i^ m/s)×(0.400k^) T) (3)

Write the formula to calculate cross product between v→ and B→ .

v→×B→=|i^j^k^a11a12a13a21a22a23|

Substitute 10.0 for a11 , 0 for a12 , 0 for a13 , 0 for a21 , 0 for a22 and 0.400 for a23 in the above expression.

v→×B→=|i^j^k^10.000000.400|=(0.400×0−0×0)i^−(10.0×0.400−0×0)j^+(10.0×0−0×0)k^=−(4.00 T⋅m/s)j^

Substitute −(4.00 T⋅m/s)j^ for v→×B→ in equation (3).

F→m=(−1.6×10−19 C)(−(4.00 T⋅m/s)j^)=6.4×10−19j^ Tm/s

The net force on the object where both fields are present is,

F→net=F→e+F→m (3)

Substitute q(v→×B→) for F→m and qE→ for F→e above expression.

F→net=qE→+q(v→×B→)

Substitute 6.4×10−19j^ Tm/s for F→m and (−4×10−19i^−8×10−19j^) N for F→e in equation (3).

F→net=(−4×10−19i^−8×10−19j^) N+(6.4×10−19j^ Tm/s)=(−4×10−19i^−1.6×10−19j^) N

The formula to calculate acceleration is,

a→=F→netme

Here,

me is mass of electron.

Substitute 9.11×10−31 kg for me and (−4×10−19i^−1.6×10−19j^) N for F→net in the above expression.

a→=(−4×10−19i^−1.6×10−19j^) N9.11×10−31 kg=(−4i^−1.6j^)×10−19 N9.11×10−31 kg=−4i^×10−19+319.11+(−1.6j^×10−19+31)9.11=−0.439i^×1012−0.176j^1012

Further solve the expression,

a→=(−4.39i^×1011−1.76j^×1011) m/s2=(−4.39i^−1.76j^)×1011 m/s2

Conclusion:

Therefore, the acceleration of electron is (−4.39i^−1.76j^)×1011m/s2

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The acceleration of electron.

Answer 12

Answer to Problem 34.4P

The acceleration of electron is (−4.39i^−1.76j^)×1011m/s2 .

Answer 13

Explanation of Solution

Given info: The uniform electric field is E→=(2.50i^+5.00j^) V/m and uniform magnetic field is B→=(0.400k^) T , and the velocity of electron is 10.0i^ m/s .