Q. 9. Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?
Answer: The Electric Generator:

An electric generator converts mechanical energy into electrical energy.
Principle: It works on the principle of electromagnetic induction. whenever there is a change in the flux through any coil, there would be the induction of current.
Working:
An electric generator, as shown consists of a rotating rectangular coil ABCD placed between the poles of a permanent magnet. The two ends of this coil are connected to the two rings R1 and R2. The inner side of the ring is insulated. The two conducting stationary brushes B1 and B2 are kept pressed separately on the rings R1 and R2, respectively. The two rings R1 and R2 are internally attached to an axle. The axle may be mechanically rotated from outside to rotate the coil inside the magnetic field. The outer ends of the two brushes are connected to the galvanometer to show the direction of the flow of current in the external circuit. When the axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Let us say the coil ABCD is rotated clockwise in the arrangement shown. By applying Fleming’s right-hand rule, the induced currents are set up in these arms along the directions AB and CD.
Thus an induced current flows in the direction ABCD. This means that the current in the external circuit flows from B2 to B1. After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both arms change, giving rise to the net induced current in the direction of DCBA. The current in the external circuit now flows from B1 to B2. Thus after every half rotation the polarity of the current in the respective arms changes. Such a current, which changes direction periodically, is called alternating current (AC). This device is called an AC generator.
To get a direct current (DC, which does not change its direction with time), a split-ring type commutator must be used. With this arrangement, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down
The function of Brushes: Brushes help in transferring the current from inside the generator to the external circuit.
Class 10 Physics NCERT Chapter 12: Higher Order Thinking Skills (HOTS) Questions
The HOTS questions in the Chapter 12: Magnetic Effects of Electric Current are carried out in order to examine analytical thinking and practical application of principles of magnetic field, electromagnetic induction and the principle of a motor. These are the questions to solve and develop critical thinking skills as well as train in preparing competitive exams.
Q1. The magnetic lines of force inside a bar magnet
a) are from north pole to south pole of the magnet
b) do not exist
c) depend upon the area of cross section of the bar magnet.
d)are from south pole to north pole of the magnet.
Answer:
The magnetic lines of force inside a bar magnet are from the south pole to the north pole of the magnet.
Q.2 A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is :
a)away from the wire
b) towards the wire
c) parallel to the wire along the current
d) parallel to the wire opposite to the current
Answer:
The direction of force on charged particle in magnetic field can also be found by fleming's left-hand rule

$\vec{F}=q(\vec{\nu} \times \vec{B})$
According to Fleming's left-hand rule, the direction of motion of the charge is towards the wire.
Hence, the answer is option (b).
Q3. The magnetic field inside a 200 turns solenoid of radius 10 cm is $2.9 \times 10^{-4} \mathrm{Tesla}$. If the solenoid carries a current of 0.29 A , then the length of the solenoid is_____$\pi \mathrm{cm}$.
Answer:
Assuming long solenoid
$
\begin{aligned}
& \mathrm{B}=\mu_0\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \\
& \ell=\frac{\mu_0 \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \\
& \ell=8 \pi \mathrm{~cm}
\end{aligned}
$
Q4. The magnetic field $B$ inside a long solenoid carrying a current of 10 A is $3.14 \times 10^{-2} \mathrm{~T}$. Find the no of turns per unit length of the Solenoid.
Answer:
For, a solenoid having n turns, then magnetic field, $B=\mu_0 n i$
$\begin{aligned}
& \Rightarrow 3.14 \times 10^{-2}=4 \pi \times 10^{-7} \times n \times 10 \\
& \Rightarrow n=\frac{3.14 \times 10^{-2}}{4 \pi \times 10^{-7} \times 10} \\
& \Rightarrow n=2500 \text { turns } / \mathrm{m}
\end{aligned}$
Hence, the answer is 2500 turns
Q5. Two long parallel wires 1 and 2 separated by a distance d carry equal currents I, as shown in Fig. The magnetic field at the point p, which is exactly midway between the two wires, is.

The magnitude of the magnetic field at a point at a distance r from a long wire carrying a current I is given by:
$\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}$
Therefore, the magnitude of the magnetic field at point $P$ is due to the current in wire 1 (since $r=d / 2$ ).
$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}}{2 \pi(\mathrm{~d} / 2)}=\frac{\mu_0 \mathrm{I}}{\pi \mathrm{~d}}$
From the right-hand thumb rule, the direction of the field is into the page. For wire 2,
$\mathrm{B}_2=\frac{\mu_0 \mathrm{I}}{\pi \mathrm{~d}} \quad \text { out of the page }$
These two fields have equal magnitude but opposite directions. Hence they cancel each other.