Question

Electron move at right angle to a magnetic field of 1.5 × 10⁻² tesla with speed of 6 × 10⁷ m/s. If the specific charge of the electron is 1.7 × 10¹¹ C/kg. The radius of circular path will be(a) 3.31 cm(b) 4.31cm(c) 1.31 cm(d) 2.35 cm

Answer 1

option D should be the answer

Answer 2

Answer: The correct answer is 2.35 cm.

Explanation:

The expression for the radius of the circular path is as follows;

$r=\frac{mv}{eB}$

Here, m is the mass of charge, v is the velocity, e is the charge of the electron and B is the magnetic field.

Rearrange the above expression.

$r=\frac{v}{\frac{e}{m}B}$

It is given in the problem that Electron move at right angle to a magnetic field of 1.5 × 10⁻² tesla with speed of 6 × 10⁷ m/s. If the specific charge of the electron is 1.7 × 10¹¹ C/kg.

Put $v=6\times 10^{7} m/s$, $\frac{e}{m}= 1.7\times 10^{11} C/kg$ and $B=1.5\times 10^{-2} T$.

$r=\frac{6\times 10^{7} }{ 1.7\times 10^{11}(1.5\times 10^{-2} )}$

$r= 2.35\times 10^{-2} m$

$r= 2.35 cm$

Therefore, the radius of the circular path is 2.35 cm.