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The force acting on a moving charge q in a magnetic field B is given by

F = qvBsinθ

Substituting the value of F,q, and sinθ (The value of sinθ = 90o ,which is equal to 1)

We get ,v = F/qBsinθ

= 3.84 x 10-14 /2 x 1.6 x 10-16 x 0.2

= 384 x 105 /2 x 16 x 2

= 6 x 105 ms-1

The speed of alpha particle is 6 x 105 ms-1

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14y ago
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2mo ago

The force on a charged particle due to a magnetic field is given by the equation F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. In this case, the force is given as 3.84 x 10^-14 N, the charge of an alpha particle is 2e (where e is the elementary charge), and the magnetic field strength is 0.2 T. Solving for v gives v = F / (qB). Plugging in the values, we get v = (3.84 x 10^-14) / (2(1.6 x 10^-19)(0.2)) = 1.2 x 10^5 m/s.

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Q: An alpha particle Helium travelling at right angles to magnetic field of 0.2 weber per m sq experience a force of 3.84 x 10-14 N.What is the speed of alpha particle?
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