Question

(17) The pulses shown in the figure are traveling through a string. Draw a snapshot graph for the string at t = 2.0 s. y (cm) at=0 1 m/s * (m) 1 m/s -1- (18) Below is a wire with current I and the current is pointing into the page. The current distribute uniformly inside the wire. Draw a graph to describe the relationship between Magnetic Field Intensity (B) and distance ® inside and outside the wire. (19) The image below shows two identical light bulbs connected to a battery and an initially uncharged capacitor. Describe the brightness of the light bulbs over time once the switch is closed, and explain your reasoning. Switch BulbA Bulb B

Answer 1

At t = 2s S = VI s = 1 X (2) = 2rrn Both waves travels a distance 2m towards each other Amplitude of the resultant wave will be given by superposition principle. 9༩） ད] s - m དག/s དེ་ ལ་དང་ ་ལ་ དེ་ 18. Magnetic field intensity B inside the wire at a distance r is given by B, irt Mohr 2« rR2

Here R is the radius of the wire , from the expression B is directly proportional to the distance from the center of the wire for inside point (r<R). Hence graph is straight line At surface r = R Hol Bin = MIR 2 x R2 = 2 x R For outside the wire magnetic field intensity B at r is given by Mo7 Bout 2 x ar Here B is inversely proportional to the distance therefore graph between B versus is a hyperbola Intensity B Brax EMOI ZAR r=R distance (8) 19. When the switch is closed capacitor is not charged, current in the circuit is maximum therefore brightness of the bulbs is maximum. As times passes capacitor starts charging and current starts decreasing exponentially. Since the resistance of the bulb is same but due to decrease in current power decreases hence brightness of the bulbs decreases with time and after a long time when capacitor is fully charged current in the circuit is zero hence bulb goes off.