Question:
A solenoid of length $50~\mathrm{cm}$ and a radius of cross-section $1\mathrm{cm}$ has 1000 turns of wire wound over it. If the current carried is 5A, the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space $\mu_{0} = 4\pi \times 10^{-7}\mathrm{T}\cdot \mathrm{m / A}$)
A solenoid of length $50~\mathrm{cm}$ and a radius of cross-section $1\mathrm{cm}$ has 1000 turns of wire wound over it. If the current carried is 5A, the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space $\mu_{0} = 4\pi \times 10^{-7}\mathrm{T}\cdot \mathrm{m / A}$)
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The magnetic field inside a solenoid is uniform and independent of the cross-sectional radius.
Updated On: Apr 8, 2026
- $0.63\times 10^{-2}\mathrm{T}$
- $1.26\times 10^{-2}\mathrm{T}$
- $2.51\times 10^{-2}\mathrm{T}$
- $6.3\mathrm{T}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The magnetic field at the centre of a long solenoid is $B = \mu_0 n I$, where $n$ is the number of turns per unit length.
Step 2: Detailed Explanation:
$n = \frac{1000}{0.5} = 2000$ turns/m.
So, $B = 4\pi \times 10^{-7} \times 2000 \times 5 = 1.26 \times 10^{-2}$ T.
Step 3: Final Answer:
The magnetic field is $1.26 \times 10^{-2}$ T.
The magnetic field at the centre of a long solenoid is $B = \mu_0 n I$, where $n$ is the number of turns per unit length.
Step 2: Detailed Explanation:
$n = \frac{1000}{0.5} = 2000$ turns/m.
So, $B = 4\pi \times 10^{-7} \times 2000 \times 5 = 1.26 \times 10^{-2}$ T.
Step 3: Final Answer:
The magnetic field is $1.26 \times 10^{-2}$ T.
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