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Chapter 30: Q. 37 (page 871)

A $20 cm 脳 20 cm$ square loop has a resistance of $0.10 惟$ . A CALC magnetic field perpendicular to the loop is $B = 4 t - 2 t^{2}$ , where $B$ is in Tesla and $t$ is in seconds. What is the current in the loop at $t = 0.0 s, t = 1.0 s$ ,and $t = 2.0 s ?$

Short Answer

Expert verified

Current in the loop at $I_{0 s}$ is $1.6 A$ .

Current in the loop at $I_{1 s}$ is $0 A$ .

Current in the loop at $I_{2 s}$ is $- 1.6 A$ .

Step by step solution

01

Formula for electric flux inside the loop

The magnetic field is expressed in terms of time in the form

$B = 4 t - 2 t^{2}$

The loop's current state is,

$I_{induced} = \frac{蔚}{R}$

$蔚 = |\frac{{诲桅}_{m}}{dt}|$

The electric flow as a function of area is,

$桅_{m} = NBA$

So,

$蔚 = NA \frac{d (BA)}{dt} = NA \frac{dB}{dt}$

02

Calculation for current in the loop att=0.0 s

Current that has been induced is,

$I_{induced} = (\frac{NA}{R}) \frac{dB}{dt}$

$= (\frac{NA}{R}) \frac{d}{dt} (4 t - 2 t^{2})$

$= (\frac{NA}{R}) (4 - 4 t)$

The area of the loop is calculated using

$A = 0.20 m 脳 0.20 m = 4 脳 10^{- 2} m^{2}$

At $t = 0.0 s$ :

Induced current is,

$I_{0 s} = (\frac{NA}{R}) (4 - 4 t)$

$= (\frac{(1) (4 脳 10^{- 2} m^{2})}{0.10 惟}) (4 - 4 (0 s))$

$= 1.6 A$

03

Calculation for current in the loop at t=1.0 s and t=2.0 s

At $t = 1.0 s$ :

Induced current is,

$I_{1 s} = (\frac{NA}{R}) (4 - 4 t)$

$= (\frac{(1) (4 脳 10^{- 2} m^{2})}{0.10 惟}) (4 - 4 (1 s))$

$= 0 A$

At $t = 2.0 s$ :

Induced current is,

$I_{2 s} = (\frac{NA}{R}) (4 - 4 t)$

$= (\frac{(1) (4 脳 10^{- 2} m^{2})}{0.10 惟}) (4 - 4 (2 s))$

$- 1.6 A$ $= (\frac{(1) (4 脳 10^{- 2} m^{2})}{0.10 惟}) (4 - 4 (1 s))$

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Most popular questions from this chapter

An $8.0 cm 脳 8.0 cm$ square loop is halfway into a magnetic CALC field perpendicular to the plane of the loop. The loop's mass is and its resistance is $0.010 惟$ . A switch is closed at $t = 0 s$ , causing the magnetic field to increase from $0$ to $1.0 T$ in $0.010 s$ .

a. What is the induced current in the square loop?

b. With what speed is the loop "kicked" away from the magnetic field?

Hint: What is the impulse on the loop?

11. a. Can you tell which of the inductors in FIGURE Q30.11 has the larger current through it? If so, which one? Explain.

b. Can you tell through which inductor the current is changing more rapidly? If so, which one? Explain.

c. If the current enters the inductor from the bottom, can you tell if the current is increasing, decreasing, or staying the same? If so, which? Explain.

FIGURE Q30.11

56. II Your camping buddy has an idea for a light to go inside your CALC tent. He happens to have a powerful (and heavy!) horseshoe magnet that he bought at a surplus store. This magnet creates a $0.20$ T field between two pole tips $10 cm$ apart. His idea is to build the hand-cranked generator shown in FIGURE P30.56. He thinks you can make enough current to fully light a $1.0 惟$ lightbulb rated at $4.0 W$ . That's not super bright, but it should be plenty of light for routine activities in the tent.

a. Find an expression for the induced current as a function of time if you turn the crank at frequency f Assume that the semicircle is at its highest point at $t = 0 s$ .

b. With what frequency will you have to turn the crank tor the maximum current to fully light the bulb? Is this feasible?

Equation 30.26 is an expression for the induced electric field CALC inside a solenoid $(r < R)$ . Find an expression for the induced electric field outside a solenoid $(r > R)$ in which the magnetic field is changing at the rate $d B / d t$ .

FIGURE EX30.5 shows a $10 cm 脳 10 cm$ square bent at a $90^{鈭�}$ angle. A uniform $0.050 T$ magnetic field points downward at a $45^{鈭�}$ angle. What is the magnetic flux through the loop?

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