Question 3: Magnetism and Matter

Question

A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the mid point between the two wires is $\_\_\_\_$ $\mu \mathrm{T}$.
$$ \left(\frac{\mu_{\mathrm{o}}}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{A}^2\right) $$

Accepted Answer

According to the Ampere's circuital Law, the magnetic field B at a distance r from an infinitely long straight wire carrying current I is :
$$ \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{~d} \overrightarrow{\mathrm{l}}=\mu_0 \mathrm{I} $$
For a circular path of radius $r$ :
$$ B(2 \pi r)=\mu_0 I \Rightarrow B=\frac{\mu_0 I}{2 \pi r} $$
Let the wires be $\mathrm{W}_1$ and $\mathrm{W}_2$ separated by $\mathrm{d}=8 \mathrm{~cm}$. The midpoint is at $\mathrm{r}=4 \mathrm{~cm}=0.04 \mathrm{~m}$ from each wire.
Wire 1 carries current along out of the page. Using the right-hand rule, the field $\mathrm{B}_1$ at the midpoint points along $y$-axis.
Wire 2 carries current along into the page. Using the right-hand rule, the field $\mathrm{B}_2$ at the midpoint also points along y -axis.
Since both fields point in the same direction, the net magnetic field $\mathrm{B}_{	ext {net }}=\mathrm{B}_1+\mathrm{B}_2$.
The current in the wires are given as $\mathrm{I}_1=\mathrm{I}_2=30 \mathrm{~A}$ and $\mathrm{r}=0.04 \mathrm{~m}$ :
$$ B_1=B_2=\frac{\mu_0 I}{2 \pi r}=2 	imes\left(\frac{\mu_0}{4 \pi}ight) \frac{I}{r} $$
$$\Rightarrow $$ $\mathrm{B}_1=\mathrm{B}_2=2 	imes 10^{-7} 	imes \frac{30}{0.04}=2 	imes 10^{-7} 	imes 750=15 	imes 10^{-5} \mathrm{~T}$
$$\Rightarrow $$ $B_{	ext {net }}=15 	imes 10^{-5}+15 	imes 10^{-5}=30 	imes 10^{-5} \mathrm{~T}$
$$\Rightarrow $$ $$ \mathrm{B}_{\mathrm{net}}=300 	imes 10^{-6} \mathrm{~T}=300 \mu \mathrm{~T} $$
Therefore, the magnitude of magnetic field at the midpoint is $300 \mu \mathrm{~T}$.
Thus, the correct option is (B).

Answer

30

Answer

300

Answer

150

Answer

0.0

Magnetism and Matter

A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the mid point between the two wires is \_\_\_\_ \mu \mathrm{T}. $ \left(\frac{\mu_{\mathrm{o}}}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{A}^2\right)

Choose the correct answer: