Question

An $\alpha$-particle experiences a force of $3.84\times {10}^{14}N$ when its moves perpendicular to magnetic field of $0.2Wb/m^2$ then speed of the $\alpha$-particle is:-

• A. $6.0\times {10}^{33}m/sec$
• B. $5.0\times {10}^{33}m/sec$
• C. $1.2\times {10}^{33}m/sec$
• D. $3.8\times {10}^{33}m/sec$

Answer 1

The correct option is A $6.0\times {10}^{33}m/sec$

$q_{\alpha }=2e=2\times 1.6\times {10}^{-19}$

$F=qv\beta$

$(3.89\times {10}^{14})=(3.2\times {10}^{-19}{)}_v(0.2)$

$V=6\times {10}^{+33}m/sec$