Question

If a straight infinitely long horizontal wire carries a current of 50 A in east-west direction, then the magnitude of the magnetic field due to the current at a vertical distance of 2 m above the wire is:

• A. 10 $\mu$T
• B. 2.5 $\mu$T
• C. 5 $\mu$T
• D. 7.5 $\mu$T

Hint:

Remember: $B$ for a long wire is $2 \times 10^{-7} \times (I/r)$.

Answer 1

The Correct Option is C

Step 1: Understand the setup.
A very long straight wire carries a current of $50\ A$. We want the magnetic field at a point $2\ m$ above the wire. The direction of the current does not change the size of the field.

Step 2: Recall the formula for a long wire.
The field at a distance $r$ from a long straight wire is $B = \dfrac{\mu_0 I}{2\pi r}$. This comes from Ampere's circuital law.

Step 3: List the known values.
Here $I = 50\ A$, $r = 2\ m$, and $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$.

Step 4: Put the numbers in.
\[ B = \frac{(4\pi\times10^{-7})(50)}{2\pi(2)} \]

Step 5: Cancel and simplify.
The $\pi$ cancels top and bottom. This leaves $B = \dfrac{(2\times10^{-7})(50)}{2} = \dfrac{100\times10^{-7}}{2}$.

Step 6: Get the value.
This equals $50\times10^{-7}\ T = 5\times10^{-6}\ T$, which is $5\ \mu T$. \[ \boxed{5~\mu T} \]