Vidaara.orgClass 12 · Physics
CodeVID-P12-02-MCF-01
Moving Charges & Magnetic Field — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Use $\mu_0=4\pi\times10^{-7}$ T m/A. Draw neat labelled diagrams wherever asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The SI unit of magnetic field is the:
- A.weber
- B.tesla
- C.henry
- D.farad
2.
Field lines around a straight wire are:
- A.straight
- B.concentric circles
- C.radial
- D.elliptical
3.
In the Biot-Savart law, $dB$ is maximum when $\theta$ equals:
- A.0 degrees
- B.90 degrees
- C.180 degrees
- D.45 degrees
4.
Field at the centre of a circular coil of $N$ turns is:
- A.$\mu_0 n I$
- B.$\frac{\mu_0 N I}{2R}$
- C.$\frac{\mu_0 I}{2\pi a}$
- D.zero
5.
The value of $\mu_0$ is:
- A.$8.85\times10^{-12}$
- B.$4\pi\times10^{-7}$ T m/A
- C.$9\times10^9$
- D.$1.6\times10^{-19}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the Biot-Savart law in vector form.
7.
Write the field inside a long solenoid and a toroid.
8.
State Ampere's circuital law.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A coil of 100 turns and radius 0.1 m carries 1 A. Find the field at its centre.
10.
Derive the field of a straight wire using Ampere's law.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State the Biot-Savart law and use it to derive the magnetic field on the axis of a circular current loop. Hence find the field at its centre.
Answer Key
Section A — Multiple Choice Questions
- (B) tesla
- (B) concentric circles
- (B) 90 degrees
- (B) $\frac{\mu_0 N I}{2R}$
- (B) $4\pi\times10^{-7}$ T m/A
Section B — Short Answer (2 marks)
- $d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$; $d\vec{B}$ is perpendicular to the plane of $d\vec{l}$ and $\hat{r}$.
- Solenoid: $B=\mu_0 n I$; toroid: $B=\frac{\mu_0 N I}{2\pi r}$.
- $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}$ for any closed loop.
Section C — Short Answer (3 marks)
- $B=\frac{\mu_0 N I}{2R}=\frac{(4\pi\times10^{-7})(100)(1)}{0.2}=6.28\times10^{-4}$ T.
- By symmetry $B(2\pi a)=\mu_0 I$, giving $B=\frac{\mu_0 I}{2\pi a}$.
Section D — Long Answer (5 marks)
- $B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ on the axis; at $x=0$, $B=\frac{\mu_0 I}{2R}$.
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