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EXMMPLE 8.1: MAGNETIC FIELD OF A SOLENOID A student takes a long straight copper wire (with insulation on it) from his phyiks tab, hews the wire on an iron rod of 50 cm , to make it solenoid of [55 (ums. He connects the ends or on w wife with the battery, which provides 1.2 A current. Find the magnetic field he profent solenoid by doing so.

Ask by Elliott Edwards. in Pakistan
Feb 22,2025

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Answer

The magnetic field of the solenoid is approximately \( 1.507964 \times 10^{-6} \) Tesla.

Solution

To find the magnetic field of the solenoid, we can use the formula for the magnetic field of a solenoid: \[ B = \mu_0 \cdot n \cdot I \] where: - \( B \) is the magnetic field strength, - \( \mu_0 \) is the magnetic constant (also known as the permeability of free space), - \( n \) is the number of turns per unit length of the solenoid, - \( I \) is the current flowing through the solenoid. Given: - The length of the iron rod (solenoid) is 50 cm = 0.5 m, - The current flowing through the solenoid is 1.2 A, - The number of turns per unit length of the solenoid is not provided, so we will assume it is 1 turn per unit length for simplicity. First, let's calculate the magnetic field strength using the given values and the formula for the magnetic field of a solenoid. Simplify the expression by following steps: - step0: Solution: \(\mu_{0}nI\) The magnetic field strength of the solenoid is given by the formula: \[ B = \mu_0 \cdot n \cdot I \] Now, we need to find the value of \( \mu_0 \) (magnetic constant) and \( n \) (number of turns per unit length) to calculate the magnetic field strength. The magnetic constant \( \mu_0 \) is approximately \( 4 \pi \times 10^{-7} \) Tm/A. Since the number of turns per unit length is not provided, we will assume it is 1 turn per unit length for simplicity. Substitute the given values into the formula to find the magnetic field strength. Calculate the value by following steps: - step0: Calculate: \(4\pi \times 10^{-7}\times 1\times 1.2\) - step1: Rewrite the expression: \(4\pi \times 10^{-7}\times 1.2\) - step2: Multiply the terms: \(4.8\pi \times 10^{-7}\) - step3: Reorder the terms: \(\pi \times 4.8\times 10^{-7}\) - step4: Reorder the terms: \(4.8\times 10^{-7}\pi \) - step5: Express with a positive exponent: \(4.8\times \frac{1}{10^{7}}\times \pi \) - step6: Multiply the terms: \(\frac{3}{6250000}\pi \) - step7: Multiply: \(\frac{3\pi }{6250000}\) The magnetic field strength of the solenoid is approximately \( 1.507964 \times 10^{-6} \) T.

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To find the magnetic field inside a solenoid, you can use the formula \( B = \mu_0 \times n \times I \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, T \cdot m/A \)), \( n \) is the number of turns per unit length, and \( I \) is the current. In this case, the student has a solenoid with 55 turns and a length of 50 cm (0.5 m). Thus, \( n = \frac{55 \, turns}{0.5 \, m} = 110 \, turns/m \). Plugging in the numbers: \[ B = (4\pi \times 10^{-7}) \times 110 \times 1.2 = 1.57 \times 10^{-4} \, T \] So, the magnetic field inside the solenoid is approximately \( 1.57 \times 10^{-4} \, T \). Ever wondered how powerful a solenoid can actually get? When you increase either the current or the number of turns, you massively boost the magnetic field strength! It’s like putting a turbocharger on your favorite toy car—watch it zoom! Just remember, safety first: high currents can heat wires; use caution when experimenting. Now, if you're curious to dive deeper into the world of electromagnetism, check out "Introduction to Electrodynamics" by David J. Griffiths. It’s a classic that gives you a solid foundation and makes students fall in love with electric and magnetic fields! Whether you're preparing for exams or just want to impress friends, this book's got you covered!

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