A charged particle of mass `m` and charge `q` travels on a circular path of radius `r` that is perpendicular to a magnetic field `B`. The time takeen by the particle to complete one revolution is

Options

A. `(2mu mq)/(B)`
B. `(2pi q^(2) B)/(m)`
C. `(2pi qB)/(m)`
D. `(2pi m)/(qB)`

Answer

To find the time taken by a charged particle of mass `m` and charge `q` to complete one revolution in a magnetic field `B`, we can follow these steps:

### Step 1: Understand the forces acting on the particle
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force required to keep it moving in a circular path. The magnetic force \( F_B \) is given by:
\[ F_B = qvB \sin \theta \]
Since the particle is moving perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \). Therefore, the magnetic force simplifies to:
\[ F_B = qvB \]

### Step 2: Set the magnetic force equal to the centripetal force
The centripetal force \( F_c \) required to keep the particle moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
Setting the magnetic force equal to the centripetal force, we have:
\[ \frac{mv^2}{r} = qvB \]

### Step 3: Solve for the radius \( r \)
From the equation above, we can rearrange it to find the radius \( r \):
\[ mv^2 = qvBr \]
Dividing both sides by \( qv \) (assuming \( v \neq 0 \)):
\[ r = \frac{mv}{qB} \]

### Step 4: Find the time period \( T \)
The time period \( T \) (the time taken to complete one revolution) is given by the formula:
\[ T = \frac{2\pi r}{v} \]
Substituting the expression for \( r \) from Step 3:
\[ T = \frac{2\pi \left( \frac{mv}{qB} \right)}{v} \]

### Step 5: Simplify the expression
The \( v \) in the numerator and denominator cancels out:
\[ T = \frac{2\pi m}{qB} \]

### Conclusion
Thus, the time taken by the particle to complete one revolution is:
\[ T = \frac{2\pi m}{qB} \]