Question

A proton moves perpendicularly to a uniform magnetic field B with a speed of 1.5 × 107 m/s and experiences an acceleration of 0.66 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction d the magnitude of the field. The elemental charge is 1.60 × 10−19 C . Answer in units of T.

Answer 1

Answer:

Magnetic field, B = 0.0045 T            

Explanation:

It is given that,

Speed of the proton, $v=1.5\times 10^7\ m/s$

Acceleration of the proton, $a=0.66\times 10^{13}\ m/s^2$

Charge on proton, $q=1.6\times 10^{-19}\ C$

The magnetic force is balanced by the force due to the acceleration of the proton as :

$qvB=ma$

$B=\dfrac{ma}{qv}$

$B=\dfrac{1.67\times 10^{-27}\ kg\times 0.66\times 10^{13}\ m/s^2}{1.6\times 10^{-19}\ C\times 1.5\times 10^7\ m/s}$

B = 0.0045 T

So, the magnitude of magnetic field on the proton is 0.0045 T. Hence, this is the required solution.