A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic force is 3.5 × 10–2 N, how fast is the charge mo…

Question

asked May 19, 2019 170k views

2 Answers

Blambert · Answer 1 · 2019-05-26T02:10:44+0000

Answer:

The charge is moving with the velocity of
$1.1*10^(4)\ m/s$ .

Step-by-step explanation:

Given that,

Charge
$q =8.4*10^(-4)\ C$

Angle = 35°

Magnetic field strength
$B=6.7*10^(-3)\ T$

Magnetic force
$F=3.5*10^(-2)\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = (F)/(qB\sin\theta)$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =(3.5*10^(-2))/(8.4*10^(-4)*6.7*10^(-3)*\sin35^(\circ))$

v =(3.5*10^(-2))/(8.4*10^(-4)*6.7*10^(-3)*0.57)

$v = 10910.36\ m/s$

$v = 1.1*10^(4)\ m/s$

Hence, The charge is moving with the velocity of
$1.1*10^(4)\ m/s$ .