Question

a point charge of 5.7 μc moves at 4.5 × 10⁵ m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram. what is the magnitude of the magnetic force acting on the charge? ○ 6.6 × 10⁻³ n ○ 4.9 × 10⁻³ n ○ 4.9 × 10³ n ○ 6.6 × 10³ n

Explanation:

Step1: Recall the formula for magnetic force on a moving charge

The magnetic force \( F \) on a moving charge \( q \) with velocity \( v \) in a magnetic field \( B \) is given by \( F = qvB\sin\theta \), where \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Step2: Identify the given values

- Charge \( q = 5.7\ \mu\text{C} = 5.7\times 10^{-6}\ \text{C} \)
- Velocity \( v = 4.5\times 10^{5}\ \text{m/s} \)
- Magnetic field \( B = 3.2\ \text{mT} = 3.2\times 10^{-3}\ \text{T} \)
- Angle \( \theta = 90^{\circ}- 37^{\circ}=53^{\circ} \)? Wait, no, looking at the diagram, the angle between velocity and the perpendicular to B (since B is horizontal left, the perpendicular is vertical). Wait, the angle between v and the perpendicular is \( 37^{\circ} \), so the angle between v and B is \( 90^{\circ}- 37^{\circ}=53^{\circ} \)? Wait, no, actually, the formula is \( \sin\theta \) where \( \theta \) is the angle between v and B. Wait, in the diagram, the velocity is at an angle of \( 37^{\circ} \) from the perpendicular to B. So the angle between v and B is \( 90^{\circ}- 37^{\circ}=53^{\circ} \)? Wait, no, maybe I got it wrong. Wait, the magnetic field is horizontal (left), velocity is at an angle of \( 37^{\circ} \) from the vertical (perpendicular to B). So the angle between v and B is \( 90^{\circ}- 37^{\circ}=53^{\circ} \)? Wait, no, actually, the component of velocity perpendicular to B is \( v\sin(90^{\circ}- 37^{\circ}) = v\cos(37^{\circ}) \)? Wait, no, let's re - examine. The magnetic force depends on the component of velocity perpendicular to the magnetic field. If the angle between velocity and the perpendicular to B is \( 37^{\circ} \), then the angle between velocity and B is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), and \( \sin\theta=\sin(53^{\circ})\approx0.8 \). Alternatively, if we consider the angle between velocity and the perpendicular to B is \( 37^{\circ} \), then the perpendicular component is \( v\cos(37^{\circ}) \)? Wait, no, let's use the formula correctly. The formula is \( F = qvB\sin\theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \). From the diagram, \( \vec{B} \) is horizontal (left), \( \vec{v} \) makes an angle of \( 37^{\circ} \) with the vertical (perpendicular to B). So the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), and \( \sin(53^{\circ})\approx0.8 \). But maybe it's easier to see that the angle between \( \vec{v} \) and the perpendicular to \( \vec{B} \) is \( 37^{\circ} \), so the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90 - 37=53^{\circ} \), and \( \sin(53^{\circ}) = 0.8 \). But let's check the values. Wait, maybe the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but let's plug in the numbers.

Wait, let's list the values again:

\( q = 5.7\times 10^{-6}\ \text{C} \)

\( v = 4.5\times 10^{5}\ \text{m/s} \)

\( B = 3.2\times 10^{-3}\ \text{T} \)

\( \theta \): The angle between \( \vec{v} \) and \( \vec{B} \). From the diagram, \( \vec{B} \) is horizontal, \( \vec{v} \) is at \( 37^{\circ} \) to the vertical (perpendicular to B), so the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), and \( \sin(53^{\circ})\approx0.8 \)

Now, calculate \( F = qvB\sin\theta \)

Substitute the values:

\( F=(5.7\times 10^{-6}\ \text{C})\times(4.5\times 10^{5}\ \text{m/s})\times(3.2\times 10^{-3}\ \text{T})\times\sin(53^{\circ}) \)

First, calculate \( 5.7\times 4.5\times 3.2\times 10^{-6 + 5- 3}\times0.8 \)

Calculate the numerical part: \( 5.7\times4.5 = 25.65 \); \( 25.65\times3.2=82.08 \); \( 82.08\times0.8 = 65.664 \)

The exponent part: \( 10^{-6 + 5- 3}=10^{-4} \)

Wait, that can't be right. Wait, maybe I made a mistake in the angle. Wait, maybe the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but maybe the diagram shows that the angle between \( \vec{v} \) and the perpendicular to \( \vec{B} \) is \( 37^{\circ} \), so the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but let's check the answer options. The options are in \( 10^{-3}\ \text{N} \) or \( 10^{3}\ \text{N} \). Let's recalculate:

Wait, maybe the angle is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but \( \sin(37^{\circ})\approx0.6 \), wait, maybe I mixed up the angle. Let's think again. The magnetic force is \( F = qvB\sin\theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \). If the velocity is at an angle of \( 37^{\circ} \) to the perpendicular of \( \vec{B} \), then the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but \( \sin(53^{\circ})\approx0.8 \), \( \sin(37^{\circ})\approx0.6 \). Wait, maybe the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but let's plug in the numbers with \( \sin(53^{\circ}) = 0.8 \):

\( q = 5.7\times 10^{-6}\ \text{C} \)

\( v = 4.5\times 10^{5}\ \text{m/s} \)

\( B = 3.2\times 10^{-3}\ \text{T} \)

\( F=(5.7\times 10^{-6})\times(4.5\times 10^{5})\times(3.2\times 10^{-3})\times0.8 \)

First, multiply the coefficients: \( 5.7\times4.5 = 25.65 \); \( 25.65\times3.2 = 82.08 \); \( 82.08\times0.8=65.664 \)

Multiply the powers of 10: \( 10^{-6}\times10^{5}\times10^{-3}=10^{-6 + 5- 3}=10^{-4} \)

So \( F = 65.664\times 10^{-4}\ \text{N}=6.5664\times 10^{-3}\ \text{N}\approx6.6\times 10^{-3}\ \text{N} \)

Wait, but let's check if the angle is \( 90^{\circ} \). No, the diagram shows an angle of \( 37^{\circ} \) from the perpendicular. Wait, maybe I made a mistake in the angle. Let's consider that the angle between \( \vec{v} \) and \( \vec{B} \) is \( 90^{\circ}- 37^{\circ}=53^{\circ} \), but let's try \( \sin(37^{\circ}) \). Wait, no, if the velocity is at \( 37^{\circ} \) to the perpendicular, then the component perpendicular to B is \( v\cos(37^{\circ}) \), so \( F = q(v\cos(37^{\circ}))B \). Let's try that:

\( F=(5.7\times 10^{-6})\times(4.5\times 10^{5})\times(3.2\times 10^{-3})\times\cos(37^{\circ}) \)

\( \cos(37^{\circ})\approx0.8 \)

So it's the same as before. Wait, \( \cos(37^{\circ})\approx0.8 \), \( \sin(53^{\circ})\approx0.8 \). So the calculation gives approximately \( 6.6\times 10^{-3}\ \text{N} \)

Step3: Verify the calculation

Let's do the calculation step by step:

1. Calculate \( qvB \) first:

\( q = 5.7\times 10^{-6}\ \text{C} \)

\( v = 4.5\times 10^{5}\ \text{m/s} \)

\( B = 3.2\times 10^{-3}\ \text{T} \)

\( qvB=(5.7\times 10^{-6})\times(4.5\times 10^{5})\times(3.2\times 10^{-3}) \)

Multiply the coefficients: \( 5.7\times4.5 = 25.65 \); \( 25.65\times3.2 = 82.08 \)

Multiply the powers of 10: \( 10^{-6}\times10^{5}\times10^{-3}=10^{-6 + 5- 3}=10^{-4} \)

So \( qvB = 82.08\times 10^{-4}\ \text{N}\cdot\text{s/C}\cdot\text{T} \) (but since \( 1\ \text{T}=1\ \text{N/(A·m)} \) and \( 1\ \text{A}=1\ \text{C/s} \), so \( 1\ \text{T}=1\ \text{N·s/(C·m)} \), and \( v \) is in m/s, so \( qvB \) has units \( \text{C}\times\text{m/s}\times\text{N·s/(C·m)}=\text{N} \))

Then multiply by \( \sin\theta \), where \( \theta = 53^{\circ} \), \( \sin(53^{\circ})\approx0.8 \)

\( F = 82.08\times 10^{-4}\times0.8=65.664\times 10^{-4}=6.5664\times 10^{-3}\approx6.6\times 10^{-3}\ \text{N} \)

Answer:

\( 6.6\times 10^{-3}\ \text{N} \) (corresponding to the first option: \( 6.6\times 10^{-3}\ \text{N} \))