Magnetism — Easy JEE physics MCQ
A long straight wire carries a current of $10$ A. The magnetic field at a perpendicular distance of $5$ cm from the wire is (Take $\mu_0 = 4\pi \times 10^{-7}$ T·m/A):
- A. $2 \times 10^{-5}$ T
- B. $4 \times 10^{-5}$ T
- C. $8 \times 10^{-5}$ T
- D. $1 \times 10^{-5}$ T
Solution
The magnetic field due to a long straight current-carrying wire at distance $r$ is:
$$B = \frac{\mu_0 I}{2\pi r}$$
Given: $I = 10$ A, $r = 5$ cm $= 0.05$ m
$$B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05}$$
$$B = \frac{4 \times 10^{-6}}{0.1} = 4 \times 10^{-5} \text{ T}$$