Question
Question: A short bar magnet allowed it to fall along the axis of the horizontal metallic ring. Starting from ...
A short bar magnet allowed it to fall along the axis of the horizontal metallic ring. Starting from rest, the distance falls by the magnet in one second may be:
A. 4.0m
B. 5.0m
C. 6.0m
D. 7.0m
Solution
Here we know that whenever a short bar magnet falls along the axis of a horizontal metallic ring, then it falls under the influence of gravity. So, we will apply the second equation of motion and solve to get the required answer.
Formula used:
s=ut+21gt2
s is the distance,
u is the initial speed,
t is the time and
g is acceleration due to gravity.
Complete answer:
Applying Newton's second law of motion.
∵s=ut+21gt2
According to the question,
Time, t=1sec
Initial velocity, u=0
Not substituting the values in above equation,
s=ut+21gt2 ⇒s=0(1)+21×10×(1)2 ⇒s=5m
So, the distance to the magnet in one second is 5m .
Hence the correct option is B .
Note:
The distance falling by the magnet in one second is less than 5m because as the magnet falls freely in the ring, flux gets increased and a field is induced whose direction is opposite to that of the bar magnet. the magnet experiences a repulsive force and hence acceleration of the magnet becomes less than g.