Answer
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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;
$\Rightarrow \vec F = q\vec v \times \vec B$ where $F$ is the magnetic force, $q$ is the charge of the particle, $\vec v$ is the velocity of the particle and $\vec B$ is the magnetic field in the region. The $ \times $ symbol in this case signifies a cross product. $W = \vec F \cdot \vec r$ where $W$ is the work done by the force $F$, and $r$ 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
$F = q\vec v \times \vec B$ where $q$ is the charge of the particle, $\vec v$ is the velocity of the particle and $\vec B$ 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
$\Rightarrow F = qvB\sin \theta $ where $\theta $ is the angle between the velocity $v$ and the field $B$. However since cross product is involved, the direction of the force $F$ is always perpendicular to both the field and the velocity, as given by the Right hand rule. Since $v$ and $r$ are always in the same direction, then $F$ is perpendicular to $r$.
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
$\Rightarrow W = \vec F \cdot \vec r$ and $r = vt$
Hence, by replacement
$\Rightarrow W = \vec F \cdot \vec vt$. Work can be written as
$\Rightarrow W = Fvt\cos \theta $ where $\theta $ is the angle between $v$ and $F$.
Since they are perpendicular, then $\theta = 90^\circ $
Hence.
$\Rightarrow W = Fvt\cos 90 = 0$
$\therefore W = 0$.
In this solution we will be using the following formula;
$\Rightarrow \vec F = q\vec v \times \vec B$ where $F$ is the magnetic force, $q$ is the charge of the particle, $\vec v$ is the velocity of the particle and $\vec B$ is the magnetic field in the region. The $ \times $ symbol in this case signifies a cross product. $W = \vec F \cdot \vec r$ where $W$ is the work done by the force $F$, and $r$ 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
$F = q\vec v \times \vec B$ where $q$ is the charge of the particle, $\vec v$ is the velocity of the particle and $\vec B$ 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
$\Rightarrow F = qvB\sin \theta $ where $\theta $ is the angle between the velocity $v$ and the field $B$. However since cross product is involved, the direction of the force $F$ is always perpendicular to both the field and the velocity, as given by the Right hand rule. Since $v$ and $r$ are always in the same direction, then $F$ is perpendicular to $r$.
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
$\Rightarrow W = \vec F \cdot \vec r$ and $r = vt$
Hence, by replacement
$\Rightarrow W = \vec F \cdot \vec vt$. Work can be written as
$\Rightarrow W = Fvt\cos \theta $ where $\theta $ is the angle between $v$ and $F$.
Since they are perpendicular, then $\theta = 90^\circ $
Hence.
$\Rightarrow W = Fvt\cos 90 = 0$
$\therefore W = 0$.
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