Question: Q. 1. (i) What is the importance of a radial magnetic field and how is it produced?
(ii) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? U[ [Delhi Comptt. I, II, III 2013]
Show Answer
Solution:
Ans. (i) Importance and production of radial magnetic field : In a radial magnetic field, magnetic torque remains maximum everywhere. $1 / 2+1 / 2$ It is produced by cylindrical pole pieces and soft iron core.
(ii) Voltmeter : A high resistance in series is required when moving coil galvanometer is used as voltmeter to make sure that a low current should travel through voltmeter without changing the potential difference then needs to be measured. $1 / 2$ Ammeter : A shunt is used when a moving coil galvanometer is used as ammeter as to make sure about not much change in total resistance of the circuit with actual current value of flowing current. $1 / 2$
[AI Q. 2. Explain giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.
U [O.D. I, II, III 2012]
Ans. (i) Try yourself, Similar to Q. 1, (ii) Short Answer Type Questions-I (AI Q. 3. A square loop of side $20 \mathrm{~cm}$ carrying current of $1 \mathrm{~A}$ is kept near an infinite long straight wire carrying a current of $2 \mathrm{~A}$ in the same plane as shown in the figure.
Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor.
A [O.D. I, II, III 2015] OR
A square shaped plane coil of area $100 \mathrm{~cm}^{2}$ of 200 turns carries a steady current of $5 \mathrm{~A}$. It is placed in a uniform magnetic field of $0.2 \mathrm{~T}$ acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle of $60^{\circ}$ with the direction of the field. In which orientation will the coil be in stable equilibrium ?
We have
$$ F=\frac{\mu_{0} I_{1} I_{2}}{2 \pi d} l $$
$\therefore$ Net force on sides $a b$ and $c d$
$=\frac{\mu_{0} 2 \times 1}{2 \pi} \times 20 \times 10^{-2}\left[\frac{1}{10 \times 10^{-2}}-\frac{1}{30 \times 10^{-2}}\right] \mathrm{N}$
$=4 \times 10^{-7} \times 20\left[\frac{20}{10 \times 30}\right] \mathrm{N}$
$=\frac{16}{3} \times 10^{-7} \mathrm{~N}$
$=5.33 \times 10^{-7} \mathrm{~N}$
This net force is directed towards the infinitely long straight wire.
Net force on sides $b c$ and $d a=$ zero.
$\therefore$ Net force on the loop $=5.33 \times 10^{-7} \mathrm{~N}$
The force is directed towards the infinitely long straight wire.
$$ \begin{aligned} & \text { OR } \ & \text { Torque }=\overrightarrow{\mu_{m}} \times \vec{B} \ &\left|\overrightarrow{\mu_{m}}\right|=n I \times A=200 \times 5 \times 100 \times 10^{-4} \text { A.m }{ }^{2} \ &=10 \mathrm{~A}^{2} \ & \text { Angle between } \overrightarrow{\mu_{m}} \text { and } \vec{B}=90^{\circ}-60^{\circ}=30^{\circ} \ & \mid \text { Torque } \mid=10 \times 0.2 \times \sin 30^{\circ} \ &=1 \mathrm{~N} . \mathrm{m} \end{aligned} $$