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QPages 249-253: Q5: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q5 Solution: A long, straight wire carrying a current of 1.0 A is placed…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the magnetic field of the wire as a scalar and simply add magnitudes, ignoring that the field circles the wire. Use the right‑hand rule to determine that at points on opposite sides of the current the wire’s field points in opposite directions, so the external field must be added on one side and subtracted on the other.

A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field

B = 1.0 \times 10^{-5} \text{ T}

pointing vertically upward figure). Find the magnitude of the resultant magnetic field at the points

P

and

Q

, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.

Step-by-Step Verified Solution

**Answer:** (a) At P:

B_{net} = 2 \times 10^{-5}

T (b) At Q:

B_{net} = 0

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

i = 1

A,

d = 0.02

m,

B_0 = 1 \times 10^{-5}

T.

B_w = \frac{\mu_0 i}{2\pi d} = \frac{4\pi \times 10^{-7} \times 1}{2\pi \times 0.02} = 1 \times 10^{-5}

T. (a) At P (one side):

B_{net} = B_w + B_0 = 2 \times 10^{-5}

T. (b) At Q (other side):

B_{net} = B_w - B_0 = 0

.

All Questions in this Chapter

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