6.) (a) A horizontal long straight wire carries a current of Amperes from left to
right. The wire is tangential to a circular loop of wire of radius which lies
immediately above the wire. The loop carries a current of in the
counter-clockwise direction. Assuming that the thicknesses of the wires are negligible,
find the magnitude and direction of the net magnetic field at the centre of the circular
loop. Explain your reasons for the direction briefly but clearly.
(b) Another long straight wire carrying a current is added so that it lies vertical and is
tangential to the circular loop on the left hand side of the loop. Find the value of which
will make the net magnetic field at the centre of the loop zero. What should be the
direction of ? Explain your answers clearly.
All wires are properly insulated and no shorting occurs.

Ask by Simpson Ford. in the United States

Mar 26,2025

Question

6.) (a) A horizontal long straight wire carries a current of Amperes from left to
right. The wire is tangential to a circular loop of wire of radius which lies
immediately above the wire. The loop carries a current of in the
counter-clockwise direction. Assuming that the thicknesses of the wires are negligible,
find the magnitude and direction of the net magnetic field at the centre of the circular
loop. Explain your reasons for the direction briefly but clearly.
(b) Another long straight wire carrying a current is added so that it lies vertical and is
tangential to the circular loop on the left hand side of the loop. Find the value of which
will make the net magnetic field at the centre of the loop zero. What should be the
direction of ? Explain your answers clearly.
All wires are properly insulated and no shorting occurs.

Ask by Simpson Ford. in the United States

Mar 26,2025

UpStudy AI · Accepted Answer

To solve this problem, we will break it down into two parts as specified.

### Part (a)

1. **Magnetic Field due to the Straight Wire**:
   The magnetic field $$ B_1 $$ at a distance $$ R $$ from a long straight wire carrying a current $$ I_1 $$ is given by the formula:
   $$
   B_1 = \frac{\mu_0 I_1}{2 \pi R}
   $$
   where $$ \mu_0 $$ is the permeability of free space, approximately $$ 4\pi \times 10^{-7} \, \text{T m/A} $$.

Given:
   - $$ I_1 = 4\pi \, \text{A} $$
   - $$ R = 0.05 \, \text{m} $$

We can calculate $$ B_1 $$.

2. **Magnetic Field due to the Circular Loop**:
   The magnetic field $$ B_2 $$ at the center of a circular loop carrying a current $$ I_2 $$ is given by:
   $$
   B_2 = \frac{\mu_0 I_2}{2R}
   $$

Given:
   - $$ I_2 = 2 \, \text{A} $$

We can calculate $$ B_2 $$.

3. **Direction of the Magnetic Fields**:
   - The magnetic field due to the straight wire (using the right-hand rule) will be directed into the page at the center of the loop.
   - The magnetic field due to the circular loop (using the right-hand rule) will be directed out of the page at the center of the loop.

4. **Net Magnetic Field**:
   The net magnetic field $$ B_{net} $$ at the center of the loop is given by:
   $$
   B_{net} = B_2 - B_1
   $$

Now, let's calculate $$ B_1 $$ and $$ B_2 $$.

### Part (b)

To find the value of $$ I_3 $$ that will make the net magnetic field at the center of the loop zero, we need to consider the magnetic field due to the vertical wire.

1. **Magnetic Field due to the Vertical Wire**:
   The magnetic field $$ B_3 $$ at the center of the loop due to the vertical wire is given by:
   $$
   B_3 = \frac{\mu_0 I_3}{2\pi R}
   $$

The direction of $$ B_3 $$ will depend on the direction of the current $$ I_3 $$.

2. **Setting the Net Magnetic Field to Zero**:
   To make the net magnetic field zero, we set:
   $$
   B_2 - B_1 - B_3 = 0
   $$
   Rearranging gives:
   $$
   B_3 = B_2 - B_1
   $$

Now, let's perform the calculations for both parts (a) and (b).

### Calculations for Part (a)

1. Calculate $$ B_1 $$:
   $$
   B_1 = \frac{\mu_0 I_1}{2 \pi R} = \frac{(4\pi \times 10^{-7}) (4\pi)}{2 \pi (0.05)}
   $$

2. Calculate $$ B_2 $$:
   $$
   B_2 = \frac{\mu_0 I_2}{2R} = \frac{(4\pi \times 10^{-7}) (2)}{2(0.05)}
   $$

### Calculations for Part (b)

1. Set up the equation for $$ I_3 $$:
   $$
   B_3 = B_2 - B_1
   $$
   $$
   \frac{\mu_0 I_3}{2\pi R} = B_2 - B_1
   $$

Now, I will perform the calculations for $$ B_1 $$ and $$ B_2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(4\times 3.14\times 10-7\right)\times 2}{\left(2\times 0.05\right)}$$
- step1: Remove the parentheses:
  $$\frac{\left(4\times 3.14\times 10-7\right)\times 2}{2\times 0.05}$$
- step2: Multiply the numbers:
  $$\frac{\left(\frac{628}{5}-7\right)\times 2}{2\times 0.05}$$
- step3: Subtract the numbers:
  $$\frac{\frac{593}{5}\times 2}{2\times 0.05}$$
- step4: Reduce the fraction:
  $$\frac{\frac{593}{5}}{0.05}$$
- step5: Convert the expressions:
  $$\frac{\frac{593}{5}}{\frac{1}{20}}$$
- step6: Multiply by the reciprocal:
  $$\frac{593}{5}\times 20$$
- step7: Reduce the numbers:
  $$593\times 4$$
- step8: Multiply the numbers:
  $$2372$$
Calculate or simplify the expression $$ (4*3.14*10**-7)*(4*3.14)/(2*3.14*0.05) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(4\times 3.14\times 10-7\right)\left(4\times 3.14\right)}{\left(2\times 3.14\times 0.05\right)}$$
- step1: Remove the parentheses:
  $$\frac{\left(4\times 3.14\times 10-7\right)\times 4\times 3.14}{2\times 3.14\times 0.05}$$
- step2: Multiply the numbers:
  $$\frac{\left(\frac{628}{5}-7\right)\times 4\times 3.14}{2\times 3.14\times 0.05}$$
- step3: Subtract the numbers:
  $$\frac{\frac{593}{5}\times 4\times 3.14}{2\times 3.14\times 0.05}$$
- step4: Reduce the fraction:
  $$\frac{\frac{593}{5}\times 4}{2\times 0.05}$$
- step5: Multiply the numbers:
  $$\frac{\frac{2372}{5}}{2\times 0.05}$$
- step6: Multiply the numbers:
  $$\frac{\frac{2372}{5}}{0.1}$$
- step7: Convert the expressions:
  $$\frac{\frac{2372}{5}}{\frac{1}{10}}$$
- step8: Multiply by the reciprocal:
  $$\frac{2372}{5}\times 10$$
- step9: Reduce the numbers:
  $$2372\times 2$$
- step10: Multiply the numbers:
  $$4744$$
### Part (a) Results

1. **Magnetic Field due to the Straight Wire**:
   $$
   B_1 = 4744 \, \text{T}
   $$

2. **Magnetic Field due to the Circular Loop**:
   $$
   B_2 = 2372 \, \text{T}
   $$

### Direction of the Magnetic Fields
- The magnetic field $$ B_1 $$ from the straight wire is directed **into the page**.
- The magnetic field $$ B_2 $$ from the circular loop is directed **out of the page**.

### Net Magnetic Field
The net magnetic field at the center of the loop is:
$$
B_{net} = B_2 - B_1 = 2372 - 4744 = -2372 \, \text{T}
$$
This indicates that the net magnetic field is directed **into the page**.

### Part (b) Calculation

To find the value of $$ I_3 $$ that will make the net magnetic field zero, we set up the equation:
$$
B_3 = B_2 - B_1
$$
Substituting the values we calculated:
$$
\frac{\mu_0 I_3}{2\pi R} = 2372 - 4744 = -2372
$$

Now, we can solve for $$ I_3 $$:
$$
I_3 = \frac{-2372 \cdot 2\pi R}{\mu_0}
$$

Substituting $$ R = 0.05 \, \text{m} $$ and $$ \mu_0 = 4\pi \times 10^{-7} $$:
$$
I_3 = \frac{-2372 \cdot 2\pi \cdot 0.05}{4\pi \times 10^{-7}}
$$

Now, I will calculate $$ I_3 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(-2372\times 2\times 3.14\times 0.05\right)}{\left(4\times 3.14\times 10-7\right)}$$
- step1: Remove the parentheses:
  $$\frac{-2372\times 2\times 3.14\times 0.05}{4\times 3.14\times 10-7}$$
- step2: Multiply the numbers:
  $$\frac{-2372\times 2\times 3.14\times 0.05}{\frac{628}{5}-7}$$
- step3: Multiply the terms:
  $$\frac{-744.808}{\frac{628}{5}-7}$$
- step4: Subtract the numbers:
  $$\frac{-744.808}{\frac{593}{5}}$$
- step5: Convert the expressions:
  $$\frac{-\frac{93101}{125}}{\frac{593}{5}}$$
- step6: Multiply by the reciprocal:
  $$-\frac{93101}{125}\times \frac{5}{593}$$
- step7: Reduce the numbers:
  $$-\frac{157}{25}\times 1$$
- step8: Multiply:
  $$-\frac{157}{25}$$
### Part (b) Result

The value of $$ I_3 $$ that will make the net magnetic field at the center of the loop zero is:
$$
I_3 \approx -6.28 \, \text{A}
$$

### Direction of $$ I_3 $$
The negative sign indicates that the direction of $$ I_3 $$ should be **in the opposite direction** to the current in the circular loop (which is counter-clockwise). Therefore, $$ I_3 $$ should be directed **downward** (or clockwise) to cancel out the magnetic field produced by the circular loop.

### Summary
- The net magnetic field at the center of the loop is directed **into the page**.
- To make the net magnetic field zero, $$ I_3 $$ should be approximately **6.28 A** directed **downward**.
At the center of the loop, the net magnetic field is directed into the page. To make the net magnetic field zero, a vertical wire carrying a current of approximately 6.28 A in the downward direction must be added.

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Solución

Part (a)

Part (b)

Calculations for Part (a)

Calculations for Part (b)

Part (a) Results

Direction of the Magnetic Fields

Net Magnetic Field

Part (b) Calculation

Part (b) Result

Direction of

Summary

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