Question

An $\alpha$ particle of mass $6.65\times {10}^{-27}kg$ travels at right angles to a magnetic field of 0.2T with a speed of $6\times {10}^{5}m/s.$ The acceleration of $\alpha$ particles will be

• A. $7.55\times {10}^{11}m{s}^{-2}$
• B. $5.77\times {10}^{11}m{s}^{-2}$
• C. $7.55\times {10}^{12}m{s}^{-2}$
• D. $5.77\times {10}^{12}m{s}^{-2}$

Answer 1

The correct option is D $5.77\times {10}^{12}m{s}^{-2}$

Magnetic force $F_b=qVB$

where, $q=2e=2\times 1.602\times {10}^{-19}$ Coulomb

V = $6\times {10}^5\frac{m}{s}$

$B=0.2T$

Now, acceleration a = $\frac{F_b}{m}$, whereas $m=6.65\times {10}^{-27}kg$.

Substitute the values and compute,

$a=\frac{2\times (1.602\times {10}^{-19})\times (6\times {10}^5)\times 0.2}{6.65\times {10}^{-27}}$
$=5.77\times {10}^{12}m{s}^{-2}$