A 5 positively charged particle is moving at 45º to the direction of magnetic field with 3 x 10 ^4 m/s speed…

1. Analyze the question

We are given:

• Charge ($q$) = $5 \mu\text{C} = 5 \times 10^{-6} \text{C}$
• Angle ($\theta$) = $45^\circ$
• Speed ($v$) = $3 \times 10^4 \text{m/s}$
• Force ($F$) = $6 \text{N}$
• We need to find the magnetic flux density ($B$).

2. Explain the concept

The force ($F$) experienced by a charged particle moving through a magnetic field is given by the Lorentz force formula:

Where:

• $F$ is the force in Newtons (N)
• $q$ is the charge in Coulombs (C)
• $v$ is the velocity in m/s
• $B$ is the magnetic flux density in Tesla (T)
• $\theta$ is the angle between the velocity and the magnetic field.

To find $B$, we rearrange the formula:

3. Show the solution

Step 1: Identify the values

• $F = 6$
• $q = 5 \times 10^{-6}$
• $v = 3 \times 10^4$
• $\theta = 45^\circ$ (Note: $\sin 45^\circ \approx 0.7071$)

Step 2: Substitute into the formula

Step 3: Simplify the denominator

• $(5 \times 10^{-6}) \times (3 \times 10^4) = 15 \times 10^{-2} = 0.15$
• $0.15 \times \sin 45^\circ = 0.15 \times 0.7071 \approx 0.106065$

Step 4: Calculate B

Rounding to two decimal places, we get 56.57 T.

4. Identify the answer

Comparing our result to the given options, the correct value is 56.57T, which corresponds to option D.

5. Tip

When solving magnetism problems, always check your units first! Convert micro-Coulombs ($\mu\text{C}$) to Coulombs (C) immediately so you don't forget the $10^{-6}$ multiplier. Also, remember that $\sin 90^\circ = 1$ (maximum force) and $\sin 0^\circ = 0$ (no force); this helps you sanity-check if your answer makes sense.