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Question: The work done by a magnetic field, on a moving charge is (A) Zero because \(\vec F\) acts paralle...

The work done by a magnetic field, on a moving charge is
(A) Zero because F\vec F acts parallel to v\vec v.
(B) Positive because F\vec F acts perpendicular to v\vec v
(C) Zero because F\vec Facts perpendicular to v\vec v
(D) Negative because F\vec Facts parallel to v\vec v

Explanation

Solution

Hint The direction of the force of the magnetic field acting on a moving charge is given by the Right hand rule. Work is said to be done if a force is applied on a body through a distance which in the direction of the force
In this solution we will be using the following formula;
F=qv×B\Rightarrow \vec F = q\vec v \times \vec B where FF is the magnetic force, qq is the charge of the particle, v\vec v is the velocity of the particle and B\vec B is the magnetic field in the region. The ×\times symbol in this case signifies a cross product. W=FrW = \vec F \cdot \vec r where WW is the work done by the force FF, and rr is the distance moved by the body while the force is being applied.

Complete step by step answer
When a body is in a magnetic field, the force exerted by the magnetic field is determined by
F=qv×BF = q\vec v \times \vec B where qq is the charge of the particle, v\vec v is the velocity of the particle and B\vec B is the magnetic field in the region. The ×\times symbol in this case signifies cross product
This can be written in magnitude only form as
F=qvBsinθ\Rightarrow F = qvB\sin \theta where θ\theta is the angle between the velocity vv and the field BB. However since cross product is involved, the direction of the force FF is always perpendicular to both the field and the velocity, as given by the Right hand rule. Since vv and rr are always in the same direction, then FF is perpendicular to rr.
Now work is only done when the distance moved by the object is, or has a component, parallel to the force.
Hence, since they are perpendicular (no component parallel) the work done is Zero.
Hence, the correct option is C.

Note
Alternatively, in mathematical terms, since work is given as
W=Fr\Rightarrow W = \vec F \cdot \vec r and r=vtr = vt
Hence, by replacement
W=Fvt\Rightarrow W = \vec F \cdot \vec vt. Work can be written as
W=Fvtcosθ\Rightarrow W = Fvt\cos \theta where θ\theta is the angle between vv and FF.
Since they are perpendicular, then θ=90\theta = 90^\circ
Hence.
W=Fvtcos90=0\Rightarrow W = Fvt\cos 90 = 0
W=0\therefore W = 0.