Question: Q. 3. (i) Show how Biot-Savart’s law can be alternatively expressed in the form of Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside a solenoid of length ’ $l$ ‘, crosssectional area ’ $A$ ’ having ’ $N$ ’ closely wound turns and carrying a steady current ’ $I$ ‘.

Draw the magnetic field lines of a finite solenoid carrying current $I$.

(ii) A straight horizontal conducting rod of length $0.45 \mathrm{~m}$ and mass $60 \mathrm{~g}$ is suspended by two vertical wires at its ends. $A$ current of $5.0 \mathrm{~A}$ is set up in the rod through the wires.

Find the magnitude and direction of the magnetic field which should be set up in order that the tension in the wire is zero.

U] A [Delhi I, II, III 2015]

Show Answer

Solution:

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

$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i $$

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 $n h$.

Hence, total enclosed current $=n h I$ Using Ampere’s circuital law

$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} n h I $$

$$ B h=\mu_{0} n h I $$

(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).

$$ I l B=m g $$

For tension to be zero,

$$ \begin{aligned} B=\frac{m g}{\mathrm{I} l} & =\frac{60 \times 10^{-3} \times 9.8}{5.0 \times 0.45} \mathrm{~T} \quad 1 / 2 \ & =0.26 \mathrm{~T} \end{aligned} $$

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 $\therefore n I A=m$

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 $\vec{\tau}$ experienced will be $\vec{\tau}=\vec{m} \times \vec{B}$

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 $F=B I l$ 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, $\quad I \propto \theta$ 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.

$$ \begin{aligned} \mathrm{I} & =\frac{C}{n B A} \theta \ \frac{\theta}{I} & =\frac{n B A}{C} \end{aligned} $$

Current Sensitivity,

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

Voltage Sensitivity, $\quad \frac{\theta}{V}=\frac{\theta}{I G}=\frac{n B A}{C G}$ 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

$$ i_{s}=\left(i-i_{g}\right) $$

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

$$ \begin{aligned} \frac{1}{R_{a}} & =\frac{1}{G}+\frac{1}{S} \ R_{a} & =\frac{G S}{G+S} \end{aligned} $$

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 $\mathrm{F}=\frac{\mu_{0}}{4 \pi} \times \frac{2 i_{1} i_{2}}{a} \times l$

  • Force between two moving charge particle $F_{m}=\frac{\mu_{0}}{4 \pi} \times \frac{q_{1} q_{2} v_{1} v_{2}}{r^{2}}$

$>\tau_{\max }=N B i A$

$>$ Current Sensitivity $=\frac{\theta}{I}=\frac{n A B}{C}$

$>$ Voltage Sensitivity $=\frac{\theta}{V}=\frac{n A B}{C G}$

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 $10 \Omega$ and that produces maximum deflection for current of $1 \mathrm{~mA}$. Find $R_{1}, R_{2}$ and $R_{3}$ that have to be used.

$$ \begin{aligned} & \text { Ans. } i_{G}\left(G+R_{1}\right)=2, R_{1}=1.99 \mathrm{k} \Omega \approx 2 \mathrm{k} \Omega \ & \begin{array}{r} i_{G}\left(G+R_{1}+R_{2}\right)=20, R_{2}=18 \mathrm{k} \Omega \quad \text { for } 2 \mathrm{~V} \text { range } \ \text { for } 20 \mathrm{~V} \text { range } \end{array} \ & i_{G}\left(G+R_{1}+R_{2}+R_{3}\right)=200, R_{3}=180 \mathrm{k} \Omega \ & \text { for } 200 \mathrm{~V} \text { range } 1 \ & \text { [CBSE Marking Scheme, } 2017] \end{aligned} $$

Detailed Answer :

From the question,

$$ \begin{aligned} & G=10 \Omega, \ & i_{G}=1 \mathrm{~mA}=10^{-3} \mathrm{~A} \end{aligned} $$

For $2 \mathrm{~V}$ range,

$$ i_{G}\left(G+R_{1}\right)=2, \mathrm{R}_{1}=1.99 \mathrm{k} \Omega \approx 2 \mathrm{k} \Omega $$

For $20 \mathrm{~V}$ range,

$$ \begin{aligned} i_{G}\left(G+R_{1}+R_{2}\right) & =20 \ R_{2} & =18 \mathrm{k} \Omega \end{aligned} $$

For $200 \mathrm{~V}$ range,

$i_{G}\left(G+R_{1}+R_{2}+R_{3}\right)=200$

$$ R_{3}=180 \mathrm{k} \Omega $$

Answering Tips

  • Sometimes students commit errors while substituting the values.


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