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:
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:
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Remember: $B$ for a long wire is $2 \times 10^{-7} \times (I/r)$.
Updated On: Jun 10, 2026
- 10 $\mu$T
- 2.5 $\mu$T
- 5 $\mu$T
- 7.5 $\mu$T
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The magnetic field ($B$) at a distance $r$ from an infinitely long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.
Step 2: Analysis
$I = 50$ A, $r = 2$ m, $\mu_0 = 4\pi \times 10^{-7}$ T$\cdot$m/A. $B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2} = \frac{2 \times 10^{-7} \times 50}{2} = 50 \times 10^{-7}$ T = $5 \mu$T.
Step 3: Conclusion
The magnetic field magnitude is 5 $\mu$T.
Final Answer: (C)
The magnetic field ($B$) at a distance $r$ from an infinitely long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$.
Step 2: Analysis
$I = 50$ A, $r = 2$ m, $\mu_0 = 4\pi \times 10^{-7}$ T$\cdot$m/A. $B = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2} = \frac{2 \times 10^{-7} \times 50}{2} = 50 \times 10^{-7}$ T = $5 \mu$T.
Step 3: Conclusion
The magnetic field magnitude is 5 $\mu$T.
Final Answer: (C)
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