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QPage 248: Q2: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Page 248: Q2 Solution: The magnetic field due to a long straight wire has been derived …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write $mu_0\epsilon_0=c^2$ instead of the correct $mu_0\epsilon_0=1/c^2$, leading to an extra factor of $c^4$ or missing $c^2$ in the denominator. Remember to solve for $mu_0$ as $mu_0=1/(\epsilon_0 c^2)$ before substituting into $B=\mu_0 i/(2\pi d)$.

The magnetic field due to a long straight wire has been derived in terms of

\mu_0

,

i

and

d

. Express this in terms of

\varepsilon_0

,

c

,

i

and

d

.

Step-by-Step Verified Solution

**Answer:**

B = \frac{i}{2\pi \epsilon_0 c^2 d}

. **Step-by-Step Explanation:** The magnetic field due to a long straight wire is

B = \frac{\mu_0 i}{2\pi d}

. Using the relation

\mu_0 \epsilon_0 = \frac{1}{c^2}

, we get

\mu_0 = \frac{1}{\epsilon_0 c^2}

. Substituting:

B = \frac{i}{2\pi \epsilon_0 c^2 d}

.

All Questions in this Chapter

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