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Given :
A cell of e.m.f 1.5 V and internal resistance 2.5 ohm is connected in series with an ammeter of resistance 0.5 ohm.
To Find :
The current in the circuit.
Solution :
We know, resistance of the ammeter is in series with the circuit.
So, total resistance is :
R = 2.5 + 0.5 ohm
R = 3 ohm
Also, e.m.f applied is 1.5 V .
Now, by ohm's law :

Therefore, the current in the circuit is 0.5 A.
Moving point charges, such as electrons, produce complicated but well known magnetic fields that depend on the charge, velocity, and acceleration of the particles. Magnetic field lines form in concentric circles around a cylindrical current-carrying conductor, such as a length of
Answer:
No, it is not magnetized.
Explanation:
Bar B does not necessarily have to be magnetized before it can be attracted to a magnet. It just has to be a magnetic material such as Iron.
If bar B were magnetized, it could either be attracted or repelled by the magnet since this would depend on the side of the pole of bar B facing it.
Since we are not given any information about bar B other than it is attracted to the magnet, it is thus not magnetized.
Answer:0.478 c
Explanation:
Given
mass of lighter Particle
mass of heavier Particle
speed of lighter particle
Let speed of heavier particle
and Momentum of the particle is given by





as momentum is conserved therefore 

