You visited us 1 times! Enjoying our articles? Unlock Full Access!

Crystal Lattice and Unit Cell

A He2+ ion ...

Question

A ${H e}^{2 +}$ ion travels at right angles to a magnetic field of 0.80 T with a velocity of $10^{5} m / s$ .If M is the magnitude of the magnetic force on the ion, find $x$ such that $x = M \times 10^{16}$ .

Open in App

Solution

$F = B q v s i n 90^{o}$
$= 0.8 \times 2 \times 1.6 \times 10^{- 19} \times 10^{5} = 2.56 \times 10^{- 14} N$

Suggest Corrections

Similar questions

α -

30^{\circ}

0.8 T

10^{5} m/s

2 \times 10^{5} m / s

0.04 T

(\frac{q}{m} = 5 \times 10^{6} C / k g)

x

x = 100 \times M

4.60 \times 10^{- 15} N

{60.0}^{o}

3.50 \times 10^{- 3} T

v

x

x = v / 10^{4}

4.0 \times 10^{5} m s^{- 1}

60^{o}

1.2 \times 10^{- X} m

Join BYJU'S Learning Program