Question

42.
A charged particle having charge 1.6 x 10-19 C
travels with a speed of 3.2 x106 ms-1 in a direction
parallel to the direction of magnetic field 0.04 T.
The force experienced by the particle is
(1) 2.0 x 10-14 N
(2) 0.2 x 10-14 N
(3) Zero
(4) 4.0 x10-14 N​

Answer 1

Answer:

The force acting on a charged particle moving in a magnetic field

B

is given by

F=q( v×B )

f = 1.6×10-¹⁹ ×10⁵

×2N

=3.2×10−¹⁴n

Answer 2

Given,

$Charge$ $q$ =  $1.6*10^{-19} C$

$Speed, v = 3.2 * 10^{6} m/s$

$Magnetic field, B = 0.04 T$

To Find,

Force experienced by the particle =?

Solution,

We can solve the question using the following steps:

The magnetic force $F$ on a charge $q$ moving at a speed $v$ in a magnetic field of strength $B$, also called Lorentz Force, is given as:

$F = q(v$×$B)$

$F = qvBsin90^{0}$

Therefore,

$F = (1.6*10^{-19} )( 3.2 * 10^{6} )(0.04)sin90^{0}$

$F = 0.2048*10^{-13} N$  {Since, $sin90^{0} = 1$}

$F = 2.0*10^{-14} N$

Hence, the force experienced by the particle is (1) 2.0 x 10-14 N.