Question

The magnetic field in a region in space (where \(x>0\) and \(y>0\) ) is given by \(\vec{B}=(x-a z) \hat{y}+(x y-b) \hat{z},\) where \(a\) and \(b\) are positive constants. An electron moving with a constant velocity, \(\vec{v}=v_{0} \hat{x},\) enters this region. What are the coordinates of the points at which the net force acting on the electron is zero?

Short answer

Answer: The net force is zero at points given by \(x = x, \; y = 0, \; z = \frac{x}{a}\). This represents a line in the positive \(x-z\) plane.

Step-by-step solution

Write down the Lorentz force equation for magnetic force

The Lorentz force equation for magnetic force is given by \(\vec{F}_{B}=q(\vec{v} \times \vec{B})\), where \(q\) is the charge of the electron, \(\vec{v}\) is the velocity, and \(\vec{B}\) is the magnetic field.

Substituting the given values for the magnetic field and velocity

We are given the magnetic field \(\vec{B}=(x-a z) \hat{y}+(x y-b) \hat{z}\) and velocity \(\vec{v}=v_{0} \hat{x}\). Substitute these values into the Lorentz force equation for magnetic force: \(\vec{F}_{B} = q(\vec{v} \times \vec{B}) = -e(v_{0} \hat{x}) \times [(x-a z) \hat{y}+(x y-b) \hat{z}]\), where \(e\) denotes the elementary charge.

Calculate the cross product \(\vec{v} \times \vec{B}\)

We now need to calculate the cross product of \(\vec{v}\) and \(\vec{B}\). We have \(\vec{v} \times \vec{B} = v_{0} \hat{x} \times [(x-a z) \hat{y}+(x y-b) \hat{z}] = -v_0[(x-a z) \hat{k} - (x y-b) \hat{j}]\).

Calculate magnetic force \(\vec{F}_{B}\)

Now we can find the magnetic force acting on the electron: \(\vec{F}_{B} = -e(v_0[(x-a z) \hat{k} - (x y-b) \hat{j}]) = (b e v_{0}) \hat{y}-(e v_{0}(x-a z)) \hat{z}\).

Find the condition for zero net force

For the net force acting on the electron to be zero, both components of \(\vec{F}_{B}\) must be zero. So, we must have \((b e v_{0}) \hat{y} = 0\) and \(-(e v_{0}(x-a z)) \hat{z} = 0\).

Solve for \(x\) and \(z\) coordinates

From \((b e v_{0}) \hat{y} = 0\), we have no constraint on coordinates \(x\) and \(z\). From \(-(e v_{0}(x-a z)) \hat{z} = 0\), we get \((x-a z) = 0\) which implies \(z = \frac{x}{a}\). Since there is no constraint on coordinates \(x\) and \(z\) from the \(\hat{y}\) component of the force, the points in the coordinate system where the net force is equal to zero will be given by: \(x = x, \; y = 0, \; z = \frac{x}{a}\). This represents a line in the positive \(x-z\) plane.