The magnetic field of the Earth at a certain location is directed vertically dow
ID: 1836109 • Letter: T
Question
The magnetic field of the Earth at a certain location is directed vertically downward and has a Magnitude of 2 times 10^minus 4T. A proton is moving horizontally towards the west in this field with a speed of 8 times 10 m/s. What is the direction and magnitude of the magnetic force the field exerts on this changes? What is the radius of the circular are followed by this proton? A 50-turn rectangular coil of dimensions 5.0 cm times 10.0 cm is dropped from a position where B = 0 to a new position where B = 0.50 T and is directetd perpendicular to the plane of yhe coil. Calculate the resulting average emf induced in the coil if the displacement occurs in 0.25s.Explanation / Answer
(9). magnetic field B = 2x10 -4 T
Speed v = 8 x10 6 m/s
Charge of proton q = 1.6 x10 -19 C
Mass of proton m = 1.67 x10 -27 kg
Direction : out of the paper i.e., along positive Z direction.
(b).In magnetic field ,Bvq = mv 2/ r
Bq = mv/r
From this radius r = mv/Bq
=(1.67 x10 -27 )(8x10 6) /(2x10 -4 )(1.6 x10 -19 )
= 417.5 m
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