A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb $\sqrt{b^2 + L^2}$ ) -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{25.0004}$ )

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

5 0

2 years ago

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai

Sliva [168]

Answer:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have: $I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

$I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}$

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode $v_1 + v_2 + v_3= 2$ and each voltage is the same v for each diode, so then:

$v = \frac{2 V}{3}= 0.667 v$

Since we have identical diodes we can use the equation:

$I_D =I= I_S e^{\frac{V_D}{V_T}}$

And replacing we have:

$I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA$

Since we know that 1 mA is drawn away from the output then the real value for I would be

$I_D = I = 3.86 mA -1 mA= 2.86 mA$

And for this case the value for $v_D$ would be:

$V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V$

And the output votage on this case would be:

$V = 3 V_D = 3 *0.660 V = 1.98 V$

And the net change in the output voltage would be:

$\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V$

5 0

2 years ago

Explain the theory Steven Reiss developed and tested.

NNADVOKAT [17]

<span>A theory of motivation by Steven Reiss, the 16 Basic Desires Theory talks about the sixteen fundamental needs, values and drives that motivate a person.</span>

6 0

2 years ago

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