Question 156: Work, Energy and Power & Circular Motion

Question

A particle of charge $1.6 \mu \mathrm{C}$ and mass $16 \mu \mathrm{~g}$ is present in a strong magnetic field of 6.28 T . The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is _________ s. $(\pi=3.14)$

Accepted Answer

To solve this problem, note that when a charged particle with mass $$m$$ and charge $$q$$ is fired perpendicular to a magnetic field of strength $$B$$, it undergoes uniform circular motion. The period (time to complete one full circle) is given by:
$$ T = \frac{2\pi m}{qB} $$
Here's how to calculate it step by step:
Convert the given quantities to SI units:
Charge: $$q = 1.6\,\mu	ext{C} = 1.6 	imes 10^{-6}\,	ext{C}$$
Mass: $$m = 16\,\mu	ext{g} = 16 	imes 10^{-9}\,	ext{kg}$$
Magnetic field: $$B = 6.28\,	ext{T}$$
Write down the period formula:
$$    T = \frac{2\pi m}{qB}    $$
Substitute the values:
$$    T = \frac{2\pi 	imes (16 	imes 10^{-9}\,	ext{kg})}{(1.6 	imes 10^{-6}\,	ext{C})(6.28\,	ext{T})}    $$
Notice that $$2\pi \approx 6.28$$, which cancels with the given magnetic field value, simplifying the expression:
$$    T = \frac{6.28 	imes 16 	imes 10^{-9}}{1.6 	imes 10^{-6} 	imes 6.28} = \frac{16 	imes 10^{-9}}{1.6 	imes 10^{-6}}    $$
Simplify the fraction:
$$    T = \frac{16}{1.6} 	imes \frac{10^{-9}}{10^{-6}} = 10 	imes 10^{-3} = 0.01\,	ext{s}    $$
Thus, the time required for the particle to return to its original location is:
$$ \boxed{0.01\,	ext{s}} $$

Work, Energy and Power & Circular Motion

A particle of charge 1.6 \mu \mathrm{C} and mass 16 \mu \mathrm{~g} is present in a strong magnetic field of 6.28 T . The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is _________ s. (\pi=3.14)

Integer / Numeric Answer Type