Travelling in magnetic fields
A particle travelling at a velocity of 100 m/s has a mass of 0.1 Kg and a charge of 2 C. The particle enters into a room of constant magnetic field of 5 Tesla through Door A. It then travels in a perfect semi-circular path (due to the influence of Magnetic Field) and exits from Door B. What is the distance, in meter, between Door A and Door B?
Assume that no Air friction acts on this particle inside the room. Note: both the doors are on the same wall (just a clarification)
The answer is 2.
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1 solution
The answer is 2m
The force exerted on a moving charged particle due to a Magnetic Field is given by F = qvb , where q = charge, v = velocity, b = Magnetic Field Strength. all the values in question have already been given in their SI Units, so no need for conversion.
Next, we also know that centripetal force (force due to moving in a circular path) = * F = (mv^{2}) / R * where m = mass, v = velocity, R = Radius of the Circular Path. We recognize that the distance between Door A and Door B is nothing by the diameter of such a circle .
Since both are forces and both are acting on this charge, we can equate these 2 formulas:
mv^{2} / R = qvb
Re-arranging to make R (radius of the circular path) the subject:
R = mv / qb; This information is straight away given in the question.
Plugging in the values will give an answer of 1 as Radius. Since the distance between the 2 doors is the diameter, distance = Radius * 2 = 2m