Step 1: Understanding the Concept:
Fleming's left-hand rule is a mnemonic used to find the direction of the force (or motion) on a current-carrying conductor in a magnetic field. This is the principle behind electric motors.
Step 2: Detailed Explanation:
To use the rule, you orient your left hand so that the thumb, index finger (forefinger), and middle finger are mutually perpendicular to each other.
\[\begin{array}{rl} \bullet & \text{The Thumb represents the direction of the Thrust or Force (Father).} \\ \bullet & \text{The Forefinger (Index finger) represents the direction of the magnetic Field (Mother).} \\ \bullet & \text{The Centre finger (Middle finger) represents the direction of the Current (Child).} \\ \end{array}\]
So, the index finger indicates the direction of the magnetic field.
Step 3: Final Answer:
According to Fleming's left-hand rule, the index finger points in the direction of the magnetic field.
Draw the pattern of the magnetic field lines for the two parallel straight conductors carrying current of same magnitude 'I' in opposite directions as shown. Show the direction of magnetic field at a point O which is equidistant from the two conductors. (Consider that the conductors are inserted normal to the plane of a rectangular cardboard.)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is: 