In uniform magnetic field, if angle between `vec(v) and vec(B) is 0^(@) lt 0 lt 90^(@)`, the path of particle is helix. Let `v_(1)` be the component of `vec(v) along vec(B) and v_(2)` be the component perpendicular to `vec(B)`. Suppose p is the pitch. T is the time period and r is the radius of helix. Then
`T = (2pim)/(qB), r = (mv_(2))/(qB), P = (v_(1))T`
Assume a charged particle of charge q and mass m is released from the origin with velocity `vec(v) = v_(0) hat(i) - v_(0) hat(k)` in a uniform magnetic field `vec(B) = -B_(0) hat(k)`.
Angle between v and B is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `120^(@)`

Explanation

To find the angle between the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\), we can follow these steps:

### Step 1: Identify the vectors
Given:
- The velocity vector \(\vec{v} = v_0 \hat{i} - v_0 \hat{k}\)
- The magnetic field vector \(\vec{B} = -B_0 \hat{k}\)

### Step 2: Calculate the magnitudes of the vectors
1. **Magnitude of \(\vec{v}\)**:
\[ |\vec{v}| = \sqrt{(v_0)^2 + (0)^2 + (-v_0)^2} = \sqrt{v_0^2 + v_0^2} = \sqrt{2v_0^2} = v_0\sqrt{2} \]

2. **Magnitude of \(\vec{B}\)**:
\[ |\vec{B}| = |-B_0| = B_0 \]

### Step 3: Calculate the dot product \(\vec{B} \cdot \vec{v}\)
The dot product is given by:
\[ \vec{B} \cdot \vec{v} = (-B_0 \hat{k}) \cdot (v_0 \hat{i} - v_0 \hat{k}) = 0 + B_0 v_0 = B_0 v_0 \]

### Step 4: Use the cosine formula
The cosine of the angle \(\theta\) between the vectors can be calculated using the formula:
\[ \cos \theta = \frac{\vec{B} \cdot \vec{v}}{|\vec{B}| |\vec{v}|} \]
Substituting the values we have:
\[ \cos \theta = \frac{B_0 v_0}{B_0 (v_0 \sqrt{2})} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the angle \(\theta\)
Now, we can find \(\theta\) using the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
This gives:
\[ \theta = 45^\circ \]

### Final Answer
The angle between \(\vec{v}\) and \(\vec{B}\) is \(45^\circ\).

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