A proton moving with a velocity of (6^i+8^j)×105ms−1 enters uniform magnetic field of induction 5×10−3^kT . The magnitude of the force acting on the proton is : (^i,^j and ^k are unit vectors forming a right handed triad)
0
8×10−16N
3×10−16N
4×10−16N
A
0
B
3×10−16N
C
4×10−16N
D
8×10−16N
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Solution
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Force on a charged particle in magnetic field is given by, ¯F=q(→V×→B)
=1.6×10−19((6^i+8^j)×5^k)×105×10−3
=1.6×10−19×10−2(−30^j+40^i)
=1.6×10−16(−3^j+4^i)
|F|=1.6×10−16√32+42
=1.6×10−16×5N
=8×10−16N
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