QJEEM19Phb5f9: JEE Main 2019 Physics - Class 12 JEEM19Phb5f9 Solution: A particle having the same charge as of electron moves in a…

Given,
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m

Charge of particle (q) = 1.6 $\times$ 10 ^{$-$ 19} C

As particle is moving in a circular path so,

${{m{v^2}} \over r} = qvB$

$\Rightarrow$   r = ${{mv} \over {qB}}$ . . . . . . . (1)

When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.

$\therefore$    F_net = 0

$\Rightarrow$    F_e = F_m

$\Rightarrow$   qE = qvB

$\Rightarrow$    E = vB  . . . . .(2)

From equation (1) and (2) we get,

r = ${m \over {qB}}\left( {{E \over B}} \right)$

= ${{mE} \over {q{B^2}}}$

$\Rightarrow$    m = ${{q{B^2}r} \over E}$

= ${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$

= 2 $\times$ 10 ^{$-$ 24} kg