First slide

Force on a charged particle moving in a magnetic field

Question

A uniform magnetic field B is acting from south to north and is of magnitude 1.5 Wb/m². If a proton having mass $= 1.7 \times 10^{- 27} kg and charge = 1.6 \times 10^{- 19} C$ moves in this field vertically downwards with energy 5 Me V, then the force acting on it will be

Moderate

A

$7.4 \times 10^{12} N$
B

$7.4 \times 10^{- 12} N$
C

$7.4 \times 10^{19} N$
D

$7.4 \times 10^{- 19} N$

Solution

$F = qvB and K = \frac{1}{2} {mv}^{2} \Rightarrow F = qB \sqrt{\frac{2 K}{m}}$
$= 1 .6 \times 10^{- 19} \times 1 .5 \sqrt{\frac{2 \times 5 \times 10^{6} \times 1 .6 \times 10^{- 19}}{1 .7 \times 10^{- 27}}}$
$= 7.344 \times 10^{- 12} N$

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