NCERT · Class 10 · Science · Magnetic Effects of Electric CurrentA long straight solenoid of length $0.5\text{ m}$ has $1000$ turns of wire wound uniformly on it. If a current of $2\text{ A}$ is passed through the solenoid, calculate the magnitude of the magnetic field inside the solenoid at its center. (Given: permeability of free space $\mu0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m}/\text{A}$).

Given Data:

• Length of the solenoid ($l$) = $0.5\text{ m}$
• Total number of turns ($N$) = $1000$
• Current flowing through the solenoid ($I$) = $2\text{ A}$
• Permeability of free space ($\mu_0$) = $4\pi \times 10^{-7} \text{ T}\cdot\text{m}/\text{A}$

Formula:\nThe magnetic field ($B$) inside a long straight solenoid at its center is given by the formula:

$$B = \mu_0 \cdot n \cdot I$$\nwhere $n$ is the number of turns per unit length, calculated as: $$n = \frac{N}{l}$|

Step-by-Step Calculation:

1. Calculate the number of turns per unit length ($n$): $$n = \frac{1000\text{ turns}}{0.5\text{ m}}$$ $$n = 2000\text{ turns/m}$|

2. Substitute the values into the magnetic field formula: $$B = (4\pi \times 10^{-7} \text{ T}\cdot\text{m}/\text{A}) \times (2000\text{ m}^{-1}) \times (2\text{ A})$$

3. Perform the multiplication: $$B = 4 \times \pi \times 10^{-7} \times 4000$$ $$B = 16000\pi \times 10^{-7} \text{ T}$$ $$B = 1.6\pi \times 10^{-3} \text{ T}$|

4. Evaluate numerically using $\pi \approx 3.1416$: $$B = 1.6 \times 3.1416 \times 10^{-3} \text{ T}$$ $$B = 5.026 \times 10^{-3} \text{ T} \text{ or } 5.03 \text{ mT}$|

Final Answer:\nThe magnitude of the magnetic field inside the solenoid at its center is $5.03 \times 10^{-3} \text{ Tesla}$ (or $5.03\text{ mT}$).