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Moving Charges and Magnetism - The Solenoid

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A solenoid is a long wire wound in the form of a helix where the neighboring turns are insulated from each other.

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For a long solenoid, the magnetic field BB is uniform and directed along the axis of the solenoid inside it.

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The magnetic field outside an ideal solenoid is practically zero because the field lines are extremely sparse.

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The direction of the magnetic field is determined by the Right-Hand Thumb Rule: if the fingers of the right hand curl in the direction of the current, the thumb points toward the North Pole of the solenoid.

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The field inside depends only on the current II and the number of turns per unit length nn, and is independent of the solenoid's radius or total length (provided it is long).

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If a soft iron core is inserted into the solenoid, the magnetic field increases significantly due to the high relative permeability μr\mu_r of the material.

📐Formulae

B=μ0nIB = \mu_0 n I

n=NLn = \frac{N}{L}

B=μ0NLIB = \mu_0 \frac{N}{L} I

Bend=12μ0nIB_{end} = \frac{1}{2} \mu_0 n I

Bmedium=μrμ0nIB_{medium} = \mu_r \mu_0 n I

💡Examples

Problem 1:

A solenoid of length 0.5 m0.5\text{ m} has a radius of 1 cm1\text{ cm} and is made up of 500500 turns. It carries a current of 5 A5\text{ A}. Calculate the magnitude of the magnetic field inside the solenoid. (Use μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7}\text{ T m/A})

Solution:

Given: L=0.5 mL = 0.5\text{ m} N=500N = 500 I=5 AI = 5\text{ A}

First, calculate the number of turns per unit length nn: n=NL=5000.5=1000 turns/mn = \frac{N}{L} = \frac{500}{0.5} = 1000\text{ turns/m}

Now, use the formula for the magnetic field inside a solenoid: B=μ0nIB = \mu_0 n I B=(4π×10−7)×1000×5B = (4\pi \times 10^{-7}) \times 1000 \times 5 B=20π×10−4 TB = 20\pi \times 10^{-4}\text{ T} B≈6.28×10−3 TB \approx 6.28 \times 10^{-3}\text{ T}

Explanation:

The magnetic field inside a long solenoid depends on the density of turns (nn) and the current (II). By calculating nn as the ratio of total turns to length, we apply Ampere's Law result to find the field strength.

Problem 2:

A solenoid has 100100 turns per cm. To produce a magnetic field of 2π×10−2 T2\pi \times 10^{-2}\text{ T} inside it, what current must be passed through it?

Solution:

Given: n=100 turns/cm=10000 turns/mn = 100\text{ turns/cm} = 10000\text{ turns/m} B=2π×10−2 TB = 2\pi \times 10^{-2}\text{ T}

Using the formula: B=μ0nIB = \mu_0 n I 2π×10−2=(4π×10−7)×10000×I2\pi \times 10^{-2} = (4\pi \times 10^{-7}) \times 10000 \times I 2π×10−2=4π×10−3×I2\pi \times 10^{-2} = 4\pi \times 10^{-3} \times I

Solving for II: I=2π×10−24π×10−3I = \frac{2\pi \times 10^{-2}}{4\pi \times 10^{-3}} I=12×101=5 AI = \frac{1}{2} \times 10^1 = 5\text{ A}

Explanation:

The number of turns per unit length must be converted to S.I. units (turns/m) before substituting into the formula. The current is then derived by rearranging the standard solenoid field equation.