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## Homework Statement

__Problem 33.12__

The loop in the figure is being pushed into the 0.20 T magnetic field at 50 m/s. The resistance of the loop is 0.1[tex]\Omega[/tex].

What is the magnitude (and direction) of the current in the loop?

__Problem 33.24__

What is the magnetic flux through the loop shown in the figure

__Problem 33.28__

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field away from vertical increases from 0.50 T to 1.50 T in 0.60 s. What is the induced emf (in mV) in the coil?

Problem 33.44

A) How big is the pushing force?

B) How much power does the pushing force supply to the wire?

C) What is the magnitude of the induced current?

D) What is the direction of the induced current?

E) How much power is dissipated in the resistor?

## Homework Equations

[tex]I = \frac{vLB}{R}[/tex]

[tex]\Phi_m = ABcos\vartheta[/tex]

[tex]E = \frac{d \Phi}{dt}[/tex]

[tex]F_{push} = \frac{v L^2 B^2}{R}[/tex]

[tex]P_{input} = F_push*v = \frac{v^2 L^2 B^2}{R}[/tex]

## The Attempt at a Solution

__Problem 33.12__

[tex]I = \frac{vLB}{R} = = \frac{(50)(0.05)(0.2)}{0.1} = 5A[/tex] and the direction is clock wise due to the right hand rule

__Problem 33.24__

[tex]A = \pi r^2[/tex]

[tex]\Phi_m = ABcos\vartheta = \pi r^2 B cos \vartheta[/tex] probably wrong

__Problem 33.28__

[tex]E = \frac{d \Phi}{dt} = \frac{d (ABcos\vartheta)}{dt} = - \pi r^2 \frac{dB}{dt}sin\vartheta[/tex]

[tex]\frac{dB}{dt} = \frac{1}{.6} = 1.67 T[/tex]

I can "ignore" the negative sign because I just need the absolute value, and the derivative of cos = -sin

*I forgot the number ot turns in the previous equation, but I added them in the next one*

[tex]N \pi r^2 \frac{dB}{dt}sin\vartheta = (100) \pi 0.1^2 (1.67) sin60 = 45mV[/tex]

__Problem 33.44__

A) [tex]F_{push} = \frac{v L^2 B^2}{R} = \frac{(0.5) (0.1)^2 (0.5)^2}{2} = 6.25 x 10^-4 N[/tex]

B) [tex]P_{input} = F_{push}*v = \frac{v^2 L^2 B^2}{R} = \frac{(0.5)^2 (0.1)^2 (0.5)^2}{2} = 3.125 x 10^4 W[/tex]

C) [tex]I = \frac{vLB}{R} = \frac{(0.5)((0.1)(0.5)}{2} = 1.25 x 10^2 A[/tex]

D) Using the right hand rule, the direction would be clockwise

E) [tex]P_dissipated = P_{input} = 3.125 x 10^4 W[/tex]

Have I made any mistakes?

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