solutions to problem set in physics 30

K .2K m= v1 2, so2

vm

=

B

e−

v

Strategy Determine the speed of the electron using its kinetic energy. Then determine the magnetic force on it.

Solution Find the speed. E

Calculate the force.18

19 1331(1.602×10 C)

9.109 10 kg

−− 2(7.2×10 J) east (0.800 T up) 10 N north−

−

⎡⎧ ⎫⎤⎨⎢⎪ ×⎥= −ev×B = − = 5.1×⎪

⎬×

⎭⎪⎥⎦⎣⎢⎪⎩

F

Strategy Solve for the speed of the proton.

Solution Find the speed.v2

= evB ,r m

so 19

727

(0.820 m)(1.602×10 C)(0.360 T) = 2.83 10× m s .1.673×10 kg

v re= Bm

−

−=

Strategy Use conservation of energy.

Solution Find the speed of the ion.

21 eV .22

∆ =K mv eV , so vm

= −∆U = =

Solve for the magnitude of the magnetic field.2

27 3

19

, so

2 1 2 1 2(12.0 u)(1.6605 kg u)(5.0 10 V)× = 0.17 T .0.21 m 1.602 10 C

evB = vm r

B m eV mVer er m r e

×10−−

mv= = = =×

Strategy Since the magnetic field in the center of the concentric circular wires is zero, the current in the smaller

loop (1) must be opposite in direction (CCW) to that of the larger loop (2). Solution Find the magnitude of the current in the smaller loop, I1.

0 2 1 12 1 1 2

12 2 1 20 , so 4.42 (8.46 A) = 6.03 A.

22 2 6.20I I r

I Irr r r r

⎞= −B B = 0 2µ µ

− 0I I1 = ⎜µ ⎛

− = =⎜ ⎟⎟⎠⎝

Thus, the current in the smaller loop is 6.03 A, CCW .

Strategy The maximum torque occurs when the field and the dipole moment are perpendicular.

Solution Compute the maximum torque.24 2 24τmax NIAB= = (9.3 10× A m⋅ )(1.0 T) = 9.3 10− −× N m⋅

1.

2.

3.

4.

5.

Solutions to Problem Set in Physics 30