Question

After being accelerated to a speed of 1.72×10⁵ m/sm/s , the particle enters a uniform magnetic field of strength 0.800 TT and travels in a circle of radius 31.0 cmcm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/qm/q.

Find the ratio m/qm/q for this particle.

Answer 1

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Answer 2

we use the formula for the radius $r$ of a charged particle's path in a magnetic field:

$$r=\frac{mv}{\mathrm{\mid }q\mathrm{\mid }B}$$

Solving for $\frac{m}{q}$$\frac{m}{q}$:

$$\frac{m}{\mathrm{\mid }q\mathrm{\mid }}=\frac{rB}{v}$$

Given:

• Speed $v=1.72\times 10^5\text{m/s}$

• Magnetic field $B=0.800\text{T}$

• Radius $r=31.0\text{cm}=0.310\text{m}$

Plugging in the values:

$$\frac{m}{\mathrm{\mid }q\mathrm{\mid }}=\frac{(0.310\text{m})(0.800\text{T})}{1.72\times 10^5\text{m/s}}=\frac{0.248}{1.72\times 10^5}\text{kg}\cdot \text{s/C}$$$$\frac{m}{\mathrm{\mid }q\mathrm{\mid }}\approx 1.44\times 10^{-6}\text{kg/C}$$