Question 705059

SCQHARD

A particle of unit mass and unit charge is moving in a magnetic field, which varies as B⃗(r⃗)=b0r⃗/r3\vec{B}(\vec{r}) = b_0 \vec{r} / r^3 (b0b_0 is a constant) over a region far away from the origin. If L⃗\vec{L} is the instantaneous angular momentum of the particle within that region, then dL⃗/dtd\vec{L}/dt is

(A)

2b0ddt(r⃗r)2 b_0 \frac{d}{dt} (\frac{\vec{r}}{r})

(B)

−b0ddt(r⃗r)-b_0 \frac{d}{dt} (\frac{\vec{r}}{r})

(C)

b0ddt(r⃗r)b_0 \frac{d}{dt} (\frac{\vec{r}}{r})

(D)

00

Detailed Solution

Force F⃗=q[v⃗×B⃗]\vec{F} = q[\vec{v} \times \vec{B}]. Torque τ⃗=r⃗×F⃗=r⃗×(v⃗×B⃗)=v⃗(r⃗⋅B⃗)−B⃗(r⃗⋅v⃗)\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times (\vec{v} \times \vec{B}) = \vec{v}(\vec{r} \cdot \vec{B}) - \vec{B}(\vec{r} \cdot \vec{v}). Given B⃗=b0r⃗r3\vec{B} = b_0 \frac{\vec{r}}{r^3}, r⃗⋅B⃗=0\vec{r} \cdot \vec{B} = 0, so τ⃗=−B⃗(r⃗⋅v⃗)=−b0r⃗r3drdt=b0ddt(r⃗r)\vec{\tau} = -\vec{B}(\vec{r} \cdot \vec{v}) = -b_0 \frac{\vec{r}}{r^3} \frac{dr}{dt} = b_0 \frac{d}{dt} (\frac{\vec{r}}{r}). Since τ⃗=dL⃗/dt\vec{\tau} = d\vec{L}/dt, the result follows.

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