Verified Solution • 2025–26 Session moderate

QPages 249-253: Q6: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q6 Solution: A long, straight wire of radius r carries a current i and is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the resultant field as only the wire's own magnetic field, forgetting that the uniform external field $\mathbf B_{ext}$ must be added vectorially (perpendicular to the azimuthal field of the wire). Remember to compute $|\mathbf B_{res}|=\sqrt{B_{wire}^2+B_{ext}^2}$, so the minimum magnitude is $B_{ext}$ (on the axis) and the maximum occurs at the surface where $B_{wire}=\mu_0 i/(2\pi r)$.

A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross section. a) At what points will the resultant magnetic field have maximum magnitude? What will be the maximum magnitude? b) What will be the minimum magnitude of the resultant magnetic field?

Step-by-Step Verified Solution

**Answer:** (a) Maximum at the surface,

B_{max} = \frac{\mu_0 i}{2\pi r}

. (b) Minimum at the axis,

B_{min} = 0

. **Step-by-Step Explanation:** For a solid wire with uniform current: - Inside:

B = \frac{\mu_0 i r}{2\pi R^2}

, increases linearly. - Outside:

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

, decreases. Maximum at the surface:

B_{max} = \frac{\mu_0 i}{2\pi R}

. Minimum at the axis:

B = 0

.

All Questions in this Chapter

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