Moving Charges and Magnetism Class 12 Case Study Questions Physics Chapter 4

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Last Updated on December 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 12 physics. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 12 physics. In this article, you will find case study questions for cbse class 12 physics chapter 4 Moving Charges and Magnetism.

Chapter	Moving Charges and Magnetism
Type of Questions	Case Study Questions
Nature of Questions	Competency Based Questions
Board	CBSE
Class	12
Subject	Physics
Unit	Unit 3 Magnetic Effects of current and Magnetism
Useful for	Class 12 Studying Students
Answers provided	Yes
Difficulty level	Mentioned
Important Link	Class 12 Physics Chapterwise Case Study

Case Study Questions on Moving Charges and Magnetism

Questions

Question 1:

Ampere’s law gives a method to calculate the magnetic field due to given current distribution.
According to this, the circulation $\oint \vec{B} \cdot d \vec{l}$ of the resultant magnetic field along a closed plane curve is equal to $\mu_0$ times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant. Ampere’s law is more useful under certain symmetrical conditions. Consider one such case of a long straight wire with circular cross-section (radius $R$ ) carrying current $I$ uniformly distributed across this cross-section.

Read the given passage carefully and give the answer of the following questions:

Q 1. What is the magnetic field at a radial distance r from the centre of the wire in the region r > R?

Q 2. What is the magnetic field at a distance r in the region r < R?

Q 3. A long straight wire of a circular cross-section (radius a) carries a steady current I and the current I is uniformly distributed across this cross-section. Plot the graph which represent the variation of magnitude of magnetic field B with distance r from the centre of the wire.

Q 4. A long straight wire of radius R carries a steady current I. The current is uniformly distributed across its cross-section. What is the ratio of magnetic field at R/2 and 2R?

Q 5. If a long straight wire in the horizontal plane carries a current of 40 A, calculate the magnitude of the field B at a point 15 cm away from the wire.

Answers

1. Magnetic field due to a long current carrying wire at $r$ is

$$
B=\frac{\mu_0}{2 \pi} \frac{l}{r} \quad(r>R)
$$

2. Let \( I’ \) be the current in the region \( r < R \). Then, \[ I' = \frac{I}{\pi R^2} \pi r^2 \quad \text{or} \quad I' = I \frac{r^2}{R^2} \] So, the magnetic field is: \[ B = \frac{\mu_0 I'}{2 \pi r} = \frac{\mu_0 \left(I \frac{r^2}{R^2}\right)}{2 \pi r} = \frac{\mu_0 I r}{2 \pi R^2} \]

3. Magnetic field due to a long straight wire of radius \( a \) carrying current \( I \) at a point distant \( r \) from the center of the wire is given as follows:

\[
\begin{aligned}
& B = \frac{\mu_0 I r}{2 \pi a^2}, \quad \text{for } r < a \\ & B = \frac{\mu_0 I}{2 \pi a}, \quad \text{for } r = a \end{aligned} \]

$$
B=\frac{\mu_0 l}{2 \pi r} \text { for } r>a
$$

The variation of magnetic field $B$ with distance $r$ from the centre of wire is shown in the figure.

4. Let the magnetic fields due to a long straight wire of radius \( R \) carrying a steady current \( I \) at a distance \( r \) from the center of the wire be:

\[
B_1 = \frac{\mu_0 I r}{2 \pi R^2} \quad (\text{for } r < R) \] \[ \text{and} \quad B_2 = \frac{\mu_0 I}{2 \pi R} \quad (\text{for } r > R)
\]

So, the magnetic field at \( r = \frac{R}{2} \) is:

\[
B_1 = \frac{\mu_0 I}{2 \pi R^2} \left(\frac{R}{2}\right) = \frac{\mu_0 I}{4 \pi R}
\]

And at \( r = 2R \):

\[
B_2 = \frac{\mu_0 I}{2 \pi (2R)} = \frac{\mu_0 I}{4 \pi R}
\]

Therefore, their corresponding ratio is:

\[
\frac{B_1}{B_2} = \frac{\left(\frac{\mu_0 I}{4 \pi R}\right)}{\left(\frac{\mu_0 I}{4 \pi R}\right)} = 1 \quad \Rightarrow \quad B_1 : B_2 = 1 : 1
\]

5.
\begin{aligned}
& \text{Given that,} \\
& I = 40 \, \mathrm{A}, \quad r = 15 \, \mathrm{cm} = 15 \times 10^{-2} \, \mathrm{m} \\
& \therefore B = \frac{\mu_0 I}{2 \pi r} = \frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 15 \times 10^{-2}} \\
& \phantom{\therefore B} = \frac{80 \times 10^{-5}}{15} = 5.34 \times 10^{-5} \, \mathrm{T}
\end{aligned}

Also check

Case study questions for other chapters of class 12 physics is given below.

• Semiconductor Electronics Class 12 Case Study Questions Physics Chapter 14
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• Atoms Class 12 Case Study Questions Physics Chapter 12
• Dual Nature of Radiation and Matter Class 12 Case Study Questions Physics Chapter 11
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• Electromagnetic Waves Class 12 Case Study Questions Physics Chapter 8
• Alternating Current Class 12 Case Study Questions Physics Chapter 7
• Electromagnetic Induction Class 12 Case Study Questions Physics Chapter 6
• Magnetism and Matter Class 12 Case Study Questions Physics Chapter 5
• Moving Charges and Magnetism Class 12 Case Study Questions Physics Chapter 4
• Current Electricity Class 12 Case Study Questions Physics Chapter 3
• Electrostatic Potential and Capacitance Class 12 Case Study Questions Physics Chapter 2
• Electric Charges and Fields Class 12 Case Study Questions Physics Chapter 1
• Case Study Questions for Class 12 Physics

We hope the given case study questions for Moving Charges and Magnetism Class 12 helps you in your learning.

Download eBooks for CBSE Class 12 Physics (Exam Special)

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Topics from which case study questions may be asked

• Concept of Magnetic Field and Oersted’s Experiment
• Biot-Savart Law and Its Application to Current-Carrying Circular Loop
• Ampere’s Law and Its Applications
• Infinitely Long Straight Wire
• Straight Solenoid
• Force on a Moving Charge in Uniform Magnetic and Electric Fields
• Force on a Current-Carrying Conductor in a Uniform Magnetic Field
• Force Between Two Parallel Current-Carrying Conductors
  • Definition of Ampere
• Torque Experienced by a Current Loop in a Uniform Magnetic Field
• Current Loop as a Magnetic Dipole and Its Magnetic Dipole Moment
• Moving Coil Galvanometer
  • Current Sensitivity
  • Conversion to Ammeter and Voltmeter

Fleming’s Left Hand Rule states that if we stretch the first finger (forefinger), the middle finger and thumb of left hand are mutually perpendicular to each other such that the first finger points in the direction of magnetic field and second finger shows the direction of moving positive charge then the thumb represents the direction of force experienced by the charged particle. The direction of force on moving negative charge is opposite to that given by the thumb.

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