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Question: A uniform magnetic field B is acting from south to north and is of magnitude 1.5\(= 1.7 \times 10 ^ ...

A uniform magnetic field B is acting from south to north and is of magnitude 1.5=1.7×10−27 kg= 1.7 \times 10 ^ { - 27 } \mathrm {~kg} and charge moves in this field vertically downwards with energy 5 MeV, then the force acting on it will be

A

B

7.4×10−12N7.4 \times 10 ^ { - 12 } N

C

D

7.4×10−19 N7.4 \times 10 ^ { - 19 } \mathrm {~N}

Answer

7.4×10−12N7.4 \times 10 ^ { - 12 } N

Explanation

Solution

F=qvBF = q v B and K=12mv2K = \frac { 1 } { 2 } m v ^ { 2 } ⇒ F=qB2kmF = q B \sqrt { \frac { 2 k } { m } }

=1.6×10−19×1.52×5×106×1.6×10−191.7×10−27= 1.6 \times 10 ^ { - 19 } \times 1.5 \sqrt { \frac { 2 \times 5 \times 10 ^ { 6 } \times 1.6 \times 10 ^ { - 19 } } { 1.7 \times 10 ^ { - 27 } } }

=7.344×10−12 N= 7.344 \times 10 ^ { - 12 } \mathrm {~N}