EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804470
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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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
Check Mark
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,

Fe=qE . (1)

Here,

q is the charge.

E is the electric field.

Fe is force due to electric field.

Substitute 1.6×1019C for q and (2.50i^+5.00j^)V/m for E in equation (1).

Fe=(1.6×1019C)((2.50i^+5.00j^)V/m)=(1.6×1019C)((2.50i^+5.00j^))N/C=(4×1019i^8×1019j^)N

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

Fm=q(v×B) (2)

Here,

B is magnetic field.

v is velocity.

Fm is force due to magnetic field.

Substitute 1.6×1019C for q , 10.0i^m/s for v and (0.400k^)T for B in equation (2).

Fm=(1.6×1019C)((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|=(0.400×00×0)i^(10.0×0.4000×0)j^+(10.0×00×0)k^=(4.00Tm/s)j^

Substitute (4.00Tm/s)j^ for v×B in equation (3).

Fm=(1.6×1019C)((4.00Tm/s)j^)=6.4×1019j^Tm/s

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

Fnet=Fe+Fm (3)

Substitute q(v×B) for Fm and qE for Fe above expression.

Fnet=qE+q(v×B)

Substitute 6.4×1019j^Tm/s for Fm and (4×1019i^8×1019j^)N for Fe in equation (3).

Fnet=(4×1019i^8×1019j^)N+(6.4×1019j^Tm/s)=(4×1019i^1.6×1019j^)N

The formula to calculate acceleration is,

a=Fnetme

Here,

me is mass of electron.

Substitute 9.11×1031kg for me and (4×1019i^1.6×1019j^)N for Fnet in the above expression.

a=(4×1019i^1.6×1019j^)N9.11×1031kg=(4i^1.6j^)×1019N9.11×1031kg=4i^×1019+319.11+(1.6j^×1019+31)9.11=0.439i^×10120.176j^1012

Further solve the expression,

a=(4.39i^×10111.76j^×1011)m/s2=(4.39i^1.76j^)×1011m/s2

Conclusion:

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

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Chapter 34 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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