A particle of mass m and charge q is in an electric and magnetic field given by `vec(E) = 2i + 3 hat(j) ,vec(B) = 4 hat(j) + 6k`.The charged particle is shifted from the origin to the point `P(x=1,Y=1)` along a straight path. The magnitude of the total work done is:

Options

A. `(0.15)q`
B. `5q`
C. `(2.5)q`
D. `(0.35)q`

Explanation

To solve the problem, we need to determine the total work done on a charged particle moving in the presence of electric and magnetic fields. Let's break it down step by step.

### Step 1: Identify the Electric and Magnetic Fields
The electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are given as:
\[ \vec{E} = 2\hat{i} + 3\hat{j} \]
\[ \vec{B} = 4\hat{j} + 6\hat{k} \]

### Step 2: Determine the Force on the Charged Particle
The force \(\vec{F}\) acting on the charged particle with charge \(q\) is given by:
\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \]
where \(\vec{v}\) is the velocity of the particle.

### Step 3: Analyze the Movement of the Particle
The particle is moved from the origin (0, 0) to the point \(P(1, 1)\). Since it is moving along a straight path and we are not given any initial velocity, we can assume that the velocity \(\vec{v}\) is zero at the starting point. Thus, the term \(q(\vec{v} \times \vec{B})\) becomes zero:
\[ \vec{F} = q\vec{E} \]

### Step 4: Calculate the Force
Substituting the electric field into the force equation:
\[ \vec{F} = q(2\hat{i} + 3\hat{j}) = 2q\hat{i} + 3q\hat{j} \]

### Step 5: Determine the Displacement Vector
The displacement \(\vec{s}\) from the origin to point \(P(1, 1)\) is:
\[ \vec{s} = 1\hat{i} + 1\hat{j} \]

### Step 6: Calculate the Work Done
The work done \(W\) is given by the dot product of the force and displacement:
\[ W = \vec{F} \cdot \vec{s} \]
Calculating the dot product:
\[ W = (2q\hat{i} + 3q\hat{j}) \cdot (1\hat{i} + 1\hat{j}) \]
\[ W = 2q(1) + 3q(1) = 2q + 3q = 5q \]

### Final Answer
The magnitude of the total work done is:
\[ W = 5q \]

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