Question

A long wire carrying a current of 5 A lies along the positive Z-axis. The magnetic field the point with position vector \(\vec{r}\) =(\(\hat{i}\)+2\(\hat{j}\)+2\(\hat{k}\)) m will be:(μ_o=4\(\pi\)×10^-7 in SI units)

• A. 2\(\sqrt{5}\)x10^-7 T
• B. 5x10^-7 T
• C. 0.33x10^-7 T
• D. 0.66x10^-7 T
• E. 7\(\sqrt{5}\) x10^-7 T

Answer 1

The Correct Option is A

Given parameters:

• Current \( I = 5 \, \text{A} \) along z-axis
• Position vector \( \vec{r} = (\hat{i} + 2\hat{j} + 2\hat{k}) \, \text{m} \)
• Permeability \( \mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} \)

 

Biot-Savart Law application: \[ \vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \hat{r}}{r^2} \] For infinite wire: \[ \vec{B} = \frac{\mu_0 I}{2\pi} \frac{\hat{z} \times \vec{r}_{perp}}{|\vec{r}_{perp}|^2} \]

Perpendicular distance calculation: \[ \vec{r}_{perp} = \hat{i} + 2\hat{j} \] \[ |\vec{r}_{perp}| = \sqrt{1^2 + 2^2} = \sqrt{5} \, \text{m} \]

Magnetic field magnitude: \[ B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(5)}{2\pi \sqrt{5}} \] \[ B = \frac{2 \times 10^{-6}}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \times 10^{-6} \] \[ B = 2\sqrt{5} \times 10^{-7} \, \text{T} \]

Thus, the correct option is (A): \( 2\sqrt{5} \times 10^{-7} \, \text{T} \).

Answer 2

1. Recall the Biot-Savart Law:

The Biot-Savart Law gives the magnetic field (dB) produced by a small current element (Idl):

\[d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}\]

where:

• μ₀ is the permeability of free space
• I is the current
• dl is the length of the current element
• r is the distance from the current element to the point where the field is being measured

2. Apply the Biot-Savart Law to an infinitely long wire:

For an infinitely long wire carrying current I along the z-axis, the magnetic field at a perpendicular distance 'a' from the wire is given by:

\[B = \frac{\mu_0 I}{2\pi a}\]

3. Determine the perpendicular distance (a):

The position vector of the point is \(\vec{r} = \hat{i} + 2\hat{j} + 2\hat{k}\). The wire lies along the z-axis. The perpendicular distance 'a' is the distance from the wire to the point in the xy-plane. We can find this by ignoring the z-component of the position vector. The x and y coordinates give (1,2), and the distance is:

\[a = \sqrt{1^2 + 2^2} = \sqrt{5} \, m\]

4. Calculate the magnetic field (B):

\[B = \frac{(4\pi \times 10^{-7} \, T \cdot m/A)(5 \, A)}{2\pi(\sqrt{5} \, m)} = \frac{10 \times 10^{-7} \, T}{\sqrt{5}} = 2\sqrt{5} \times 10^{-7} \, T\]

Final Answer: The final answer is \(\boxed{A}\)