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2 years ago

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Physics

2 answers:

Alexxx [7]2 years ago

5 0

The correct answer is A) 1.6 x 10-1 N

Gekata [30.6K]2 years ago

4 0

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = $6.5*10^{4}\ m/s$

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity $[\theta]$ = 15 degree.

We are asked to calculate the magnetic force experienced by the charged particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force $\vec F=q(\ \vec V \times \vec B)$

i.e F = $=qVBsin\theta$

$=6.8*6.5*10^{4}*1.4*sin15$

$=61.88*10^{4}sin15$

$=61.88*10^4*0.2588\ N$

$=16.016*10^4\ N$

$=1.6*10^5\ N$

Hence, the force experienced by the charged particle is C i.e $1.6*10^5 N$

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