Solution:

Ans. (i) A current carrying surface with small line elements of length $dl$ where tangential components. of magnetic field and sum of elements B.dl in an integral form be expressed as :

$$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}i$$

This form is known as Ampere’s circuital law. 1/2

$1/2$

Let ’ $n$ ’ be the number of turns per unit length. Then total number of turns in the length ’ $h$ ’ is $nh$.

Hence, total enclosed current $=nhI$ Using Ampere’s circuital law

$$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}nhI$$

$$Bh={\mu }_{0}nhI$$

(ii)

As per the given figure, magnetic field must be vertically inwards, to make tension zero, (If a student shows current in opposite direction the magnetic field should be set up vertically upwards).

$$IlB=mg$$

For tension to be zero,

$$\begin{array}{rlr}B=\frac{mg}{\mathrm{I}l}& =\frac{60\times {10}^{-3}\times 9.8}{5.0\times 0.45}\mathrm{T}1/2& =0.26\mathrm{T}\end{array}$$

TOPIC-3

Torque and Galvanometer

Revision Notes

Torque experienced by a current loop in uniform magnetic field

In a rectangular loop of length $l$, breadth $b$ with current $I$ flowing through it in a uniform magnetic field of induction $B$ where angle $\theta$ is between the normal and in direction of magnetic field, then the torque experienced will be :

where, $n=$ number of turns in the coil $\therefore nIA=m$

Further,

Torque will be maximum when the coil is parallel to magnetic field and will be zero when coil is perpendicular to magnetic field.

In vector notation, torque $\overrightarrow{\tau }$ experienced will be $\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}$

Moving coil galvanometer

It is an instrument used for detection and measurement of small electric currents

In this, when a current carrying coil is suspended in uniform magnetic field, it experiences a torque which rotates the coil.

The force experienced by each side of the galvanometer will be $F=BIl$ which are opposite in direction.

Opposite and equal forces form the cquple which generates deflecting torque on the coil having number of turns $n$ is given as :

In moving coil galvanometer, current in the coil will be directly proportional to the angle of the deflection of the coil, $I\propto \theta$ i.e., where, $\theta$ is the angle of deflection.

Current sensitivity of galvanometer

Current sensitivity of galvanometer is the deflection produced when unit current passes through the galvanometer. A galvanometer is said to be sensitive if it produces large deflection for a small current.

$$\begin{array}{rlr}\mathrm{I}& =\frac{C}{nBA}\theta \frac{\theta }{I}& =\frac{nBA}{C}\end{array}$$

Current Sensitivity,

Voltage sensitivity of galvanometer is the deflection per unit voltage given as

Voltage Sensitivity, $\frac{\theta }{V}=\frac{\theta }{IG}=\frac{nBA}{CG}$ where, $G=$ galvanometer resistance, $C=$ torsional constant.

Increase in sensitivity of moving coil galvanometer depends on :

(i) number of turns $n$ (ii) magnetic field $B$ (iii) area of coil $A$ and (iv) torsional constant.

Conversion of galvanometer into ammeter

Galvanometer can be converted into ammeter by connecting a low resistance known as shunt in parallel with the galvanometer coil.

If $I_g$ being the maximum current with full scale deflection that passes through galvanometer, then current through shunt resistance will be

$$i_s=(i-i_g)$$

where, $G=$ Galvanometer resistance, $S=$ Shunt resistance and $i=$ Current in circuit

D Now effective resistance of ammeter will be :

Conversion of galvanometer into voltmeter

$$\begin{array}{rlr}\frac{1}{R_a}& =\frac{1}{G}+\frac{1}{S}{R}_a& =\frac{GS}{G+S}\end{array}$$

Voltmeter measures the potential difference between the two ends of a current carrying conductor.

Galvanometer can be converted to voltmeter by connecting high resistance in series with galvanometer coil.

As resistance $R$ is connected in series with galvanometer, current through the galvanometer will be,

• Effective resistance of voltmeter is

where, $R_v$ is very large making the voltmeter to onnect in parallel since it can draw less current from the circuit.

Key Formulae

Force between two parallel wires $\mathrm{F}=\frac{{\mu }_0}{4\pi }\times \frac{2i_1{i}_2}{a}\times l$

• Force between two moving charge particle $F_m=\frac{{\mu }_0}{4\pi }\times \frac{q_1{q}_2{v}_1{v}_2}{r^2}$

$>{\tau }_{max}=NBiA$

$>$ Current Sensitivity $=\frac{\theta }{I}=\frac{nAB}{C}$

$>$ Voltage Sensitivity $=\frac{\theta }{V}=\frac{nAB}{CG}$

f. Very Short Answer Type Questions

[II Q. 1. A multi-range voltmeter can be constructed by using the galvanometer circuit as shown. We want to construct a voltmeter that can measure $2\mathrm{V},20\mathrm{V}$ and $200\mathrm{V}$ using a galvanometer of resistance $10\mathrm{\Omega }$ and that produces maximum deflection for current of $1\mathrm{mA}$. Find $R_1,R_2$ and $R_3$ that have to be used.

$$\begin{array}{rlrlrl}& \text{ Ans. }{i}_G(G+R_1)=2,R_1=1.99\mathrm{k}\mathrm{\Omega }\approx 2\mathrm{k}\mathrm{\Omega }& \begin{array}{c}{i}_G(G+R_1+R_2)=20,R_2=18\mathrm{k}\mathrm{\Omega }\text{ for }2\mathrm{V}\text{ range }\text{ for }20\mathrm{V}\text{ range }\end{array}& i_G(G+R_1+R_2+R_3)=200,R_3=180\mathrm{k}\mathrm{\Omega }& \text{ for }200\mathrm{V}\text{ range }1& \text{ [CBSE Marking Scheme, }2017]\end{array}$$

Detailed Answer :

From the question,

$$\begin{array}{rlr}& G=10\mathrm{\Omega },& i_G=1\mathrm{mA}={10}^{-3}\mathrm{A}\end{array}$$

For $2\mathrm{V}$ range,

$$i_G(G+R_1)=2,{\mathrm{R}}_1=1.99\mathrm{k}\mathrm{\Omega }\approx 2\mathrm{k}\mathrm{\Omega }$$

For $20\mathrm{V}$ range,

$$\begin{array}{rlr}{i}_G(G+R_1+R_2)& =20R_2& =18\mathrm{k}\mathrm{\Omega }\end{array}$$

For $200\mathrm{V}$ range,

$i_G(G+R_1+R_2+R_3)=200$

$$R_3=180\mathrm{k}\mathrm{\Omega }$$

Answering Tips

• Sometimes students commit errors while substituting the values.