Chapter 30: Q. 85 (page 875)

85. III A 2.0-cm-diameter solenoid is wrapped with 1000 turns per CALC meter. 0.50 cm from the axis, the strength of an induced electric field is $5.0 \times 10^{- 4} V / m$ . What is the rate $d I / d t$ with which the current through the solenoid is changing?

Short Answer

Expert verified

The change in current over the change in time is $160 A / s$

Step by step solution

Step 1. given information

Diameter of the solenoid, $D = 2 cm = 0.02 m$ , number of turns per meter is $n = 1000$ , distance $d = 0.5 cm = 0.005 m$ from the axis, the strength of induced electric field is $E = 5 \times 10^{- 4} V / m .$

Step 2. Explanation

The equation for change in current over the change in time, $\frac{d I}{d t}$ .

$\frac{d I}{d t} = \frac{E 2 l}{r μ_{0} N}$

Convert from to meters.

$r = (0.50 cm) (\frac{1 m}{100 cm}) = 0.0050 m$

Calculate the rate of change of change of current with respect to change in time.

$\frac{d I}{d t} = \frac{E 2 l}{r μ_{0} N}$

Substitute $5.0 \times 10^{- 4} V / m$ for $E, 0.0050 m$ for $r$ , and 1000 for turns per length $\frac{N}{l}$ .

$\begin{array}{l} \frac{d I}{d t} = \frac{2 E}{r μ_{0} (\frac{N}{l})} \\ = \frac{2 (5.0 \times 10^{- 4} V / m)}{(0.0050 m) (4 π \times 10^{- 7} Tm / A) (1000 m^{- 1})} \\ = 160 A / s \end{array}$

Thus, the change in current over the change in time is $160 A / s$ .

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85. III A 2.0-cm-diameter solenoid is wrapped with 1000 turns per CALC meter. 0.50 cm from the axis, the strength of an induced electric field is $5.0 \times 10^{- 4} V / m$ . What is the rate $d I / d t$ with which the current through the solenoid is changing?

Short Answer

Step by step solution

Step 1. given information

Step 2. Explanation

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85. III A 2.0-cm-diameter solenoid is wrapped with 1000 turns per CALC meter. 0.50 cm from the axis, the strength of an induced electric field is5.0×10−4V/m . What is the rate dI/dtwith which the current through the solenoid is changing?

Short Answer

Step by step solution

Step 1. given information

Step 2. Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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85. III A 2.0-cm-diameter solenoid is wrapped with 1000 turns per CALC meter. 0.50 cm from the axis, the strength of an induced electric field is $5.0 \times 10^{- 4} V / m$ . What is the rate $d I / d t$ with which the current through the solenoid is changing?