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A solenoid of length 0.5 m and radius 1 cm has 500 turns carrying a current of 5 A. The magnitude of the field inside it is:
A3.14 × 10⁻³ T
B6.28 × 10⁻⁴ T
C6.28 × 10⁻³ T
D1.26 × 10⁻² T
Answer & Solution
Correct answer: C. 6.28 × 10⁻³ T
1. Turns per unit length: $n = \dfrac{500}{0.5} = 1000$ m⁻¹.
2. Inside field: $B = \mu_0 n I$.
3. Substitute $\mu_0 = 4\pi\times10^{-7}$, $n = 10^3$, $I = 5$.
4. $B = 4\pi\times10^{-7}\times10^{3}\times5 = 2\pi\times10^{-3}$.
5. $B = 6.28\times10^{-3}$ T.
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.153_
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