HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q16 Solution: A long, straight wire carries a current . Let be the magnetic…

QPages 249-253: Q16: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q16 Solution: A long, straight wire carries a current . Let be the magnetic…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently misapply the finite wire magnetic field formula by omitting the square root term or using the wrong geometric factor, leading to incorrect algebraic setup. The correct expression for a segment of length $l$ centered at distance $d$ is $B_2 = \frac{\mu_0 i}{4\pi d} \frac{l}{\sqrt{d^2 + l^2/4}}$, which must be carefully equated to $0.99 B_1$ to solve for the ratio.

A long, straight wire carries a current

i

. Let

B_1

be the magnetic field at a point

P

at a distance

d

from the wire. Consider a section of length

l

of this wire such that the point P lies on a perpendicular bisector of the section. Let

B_2

be the magnetic field at this point due to this section only. Find the value of

d/l

so that

B_2

differs from

B_1

by 1%.

Step-by-Step Verified Solution

**Answer:**

d/l = 14.1

. **Step-by-Step Explanation:**

B_1 = \frac{\mu_0 i}{2\pi d}

(infinite wire).

B_2 = \frac{\mu_0 i l}{4\pi d\sqrt{d^2 + l^2/4}}

(finite section). Given

B_2 = 0.99 B_1

\frac{l}{2\sqrt{d^2 + l^2/4}} = 0.99 \Rightarrow \frac{l^2}{4(d^2 + l^2/4)} = 0.98

. Solving:

d/l \approx 14.1

QPages 249-253: Q16: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q16 Solution: A long, straight wire carries a current . Let be the magnetic…

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