JEE PYQ: Magnetic Effects Of Current Question 26
Question 26 - 2020 (08 Jan 2020 Shift 1)
Proton with kinetic energy of $1$ MeV moves from south to north. It gets an acceleration of $10^{12}$ m/s$^2$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27}$ kg)
(1) $0.71$ mT
(2) $7.1$ mT
(3) $0.071$ mT
(4) $71$ mT
Show Answer
Answer: (1) $0.71$ mT
Solution
$a = \frac{qvB}{m}$. $v = \sqrt{\frac{2KE}{m}}$. Solving: $B = \frac{1}{\sqrt{2}} \times 10^{-3}$ T $\approx 0.71$ mT.
BETA
