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Chapter 26: Problem 42

A particle carrying a \(50-\mu \mathrm{C}\) charge moves with velocity \(\vec{v}=5.0 \hat{\imath}+3.2 \hat{k} \mathrm{m} / \mathrm{s}\) through a magnetic field \(\vec{B}=9.4 \hat{\imath}+6.7 \hat{\jmath} \mathrm{T}\) (a) Find the magnetic force on the particle. (b) Form the dot products \(\vec{F} \cdot \vec{v}\) and \(\vec{F} \cdot \vec{B}\) to show explicitly that the force is perpendicular to both \(\vec{v}\) and \(\vec{B}\).

Short Answer

Expert verified
The magnetic force \(F\) acting on the particle is directly proportional to the cross product of its velocity and the magnetic field. Thus, \(F\) is perpendicular to both the velocity and the magnetic field, as demonstrated by the zero value of the dot products \(F \cdot v\) and \(F \cdot B\).

Step by step solution

01

Compute Magnetic Force

Calculate the vector cross product of the velocity (\(v\)) and the magnetic field (\(B\)). Multiply the result by the charge of the particle (q). This will give the magnetic force. \(F = q(v \times B)\).
02

Compute the dot product of Magnetic Force and velocity

Use the formula of the dot product to calculate the dot product of the magnetic force \(F\) found in step 1 and the velocity vector \(v\). The dot product is defined as \(F \cdot v = |F||v| \cos(\theta)\), where \(\theta\) is the angle between \(F\) and \(v\). If \(F\) and \(v\) are perpendicular, the value of the dot product is zero.
03

Compute the dot product of Magnetic Force and Magnetic Field

Use the formula of the dot product to calculate the dot product of the magnetic force \(F\) found in step 1 and the magnetic field vector \(B\). The dot product is defined as \(F \cdot B = |F||B| \cos(\theta)\), where \(\theta\) is the angle between \(F\) and \(B\). If \(F\) and \(B\) are perpendicular, the value of the dot product is zero.

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