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QPages 249-253: Q7: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q7 Solution: A long, straight wire carrying a current of 30 A is placed in…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the magnetic field generated by a straight current runs along the wire and therefore add it directly to an external field that is parallel to the current; in reality the wire’s field is azimuthal (circles the wire) and is always perpendicular to any uniform field aligned with the wire, so the magnitudes must be combined using the Pythagorean sum.

A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of

4.0 \times 10^{-4} \text{ T}

parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire.

Step-by-Step Verified Solution

**Answer:**

B_{net} = 5 \times 10^{-4}

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

i = 30

A,

d = 0.02

m,

B_0 = 4 \times 10^{-4}

T.

B_w = \frac{\mu_0 i}{2\pi d} = \frac{4\pi \times 10^{-7} \times 30}{2\pi \times 0.02} = 3 \times 10^{-4}

T. Since

B_0

and

B_w

are perpendicular:

B_{net} = \sqrt{(4 \times 10^{-4})^2 + (3 \times 10^{-4})^2} = 5 \times 10^{-4}

T.

All Questions in this Chapter

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