Question

a charge q moves with a velocity Vinto magnetic field 'B' perpendicularly,then the force exerted on it is:
A)qVB cos 90°
B)qVB tan 90°
C)qVB cot 90°
D)qVB​

Answer 1

The expression for force experienced by a charge is......

F = qvB sin theta

Here .... theta = 90°

Therefore....sin theta = 1

....so....... F = qvB...

Answer 2

Force exerted on the charged particle is qVB .

Given :

Charge on particle , q .

Velocity of charged particle , V .

Magnetic field , B .

It is also given that the angle between velocity ( V ) and magnetic field ( B ) is $90^o$ .

We need to find the force on charged particle .

We know ,

$\vec{F}=q(\vec{V}\times \vec{B} )\\\\F=qVBsin\ 90^o\\\\F=qVB$ { $sin 90^o=1$ }

Therefore , option D) is correct .

Hence , this is the required solution .

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Electromagnetism

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