A proton, a deuteron and an $\alpha$ particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is _________ and their speed is _______, in the ratio.
Solution
F = qvB = q${p \over m}$B<br><br>F $\propto$ ${q \over m}$ [as p, B are const.]<br><br>$\therefore$ F<sub>1</sub> : F<sub>2</sub> : F<sub>3</sub><br><br>= ${e \over m}:{e \over {2m}}:{{2e} \over {4m}}$<br><br>= 2 : 1 : 1<br><br>And v<sub>1</sub> : v<sub>2</sub> : v<sub>3</sub><br><br>= ${p \over m}:{p \over {2m}}:{p \over {4m}}$<br><br>= 4 : 2 : 1
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Magnetic Effects of Current are available — start with the harder ones if your accuracy is >70%.