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184_notes:examples:week10_force_on_charge [2017/10/29 21:00] – [Solution] tallpaul | 184_notes:examples:week10_force_on_charge [2017/10/29 21:54] – [Solution] tallpaul | ||
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- | The trickiest part of finding magnetic force is the cross-product. | + | Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$. |
+ | |||
+ | The trickiest part of finding magnetic force is the cross-product. | ||
+ | |||
+ | \begin{align*} | ||
+ | \vec{v} &= \langle 10, 0, 0 \rangle \text{ m/s} \\ | ||
+ | \vec{B} &= \langle 0, 4\cdot 10^{-4}, 0 \rangle \text{ T} \\ | ||
+ | \vec{v} \times \vec{B} &= \langle v_y B_z - v_z B_y, v_z B_x - v_x B_z, v_x B_y - v_y B_x \rangle \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | Alternatively, | ||
+ | |||
+ | $$\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/ | ||
+ | |||
+ | We get the same answer with both methods. Now, for the force calculation: | ||
+ | |||
+ | $$\vec{F} = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ | ||
+ | |||
+ | Notice that the $\sin 90^{\text{o}}$ term evaluates to $1$. This is the maximum value it can be, and we get this value because the velocity and the B-field were exactly perpendicular. Any other orientation would have yielded a weaker magnetic force. In fact, in the second case, when the velocity is parallel |