Question

(a) A long straight wire carries a steady current $I$ along the positive $y$-axis in a three-dimensional coordinate system. A particle of charge $+Q$ is moving with a velocity $\vec{v} = v_0\hat{i}$. Find the magnitude and direction of the magnetic force experienced by the particle at the instant it is at $(x, 0, 0)$.

(b) A galvanometer of resistance $100\,\Omega$ shows full scale deflection for a current of $5\,\text{mA}$. Explain how you will convert it into a voltmeter of range $(0 - 1\,\text{V})$. Obtain the value of the resistance required.

Hint:

Find B of the wire at the point, then use F = Q v x B. For a voltmeter put a series resistance R = V/Ig - G.

Answer 1

Step 1: Understanding the Concept:
A long straight current produces a magnetic field that circles the wire. A moving charge in this field feels the Lorentz force \(\vec{F} = Q\,\vec{v} \times \vec{B}\).

Step 2: Part (a): Field at (x, 0, 0):
Take the point on the positive \(x\) side (\(x > 0\)). Magnitude at distance \(x\) from the wire: \(B = \dfrac{\mu_0 I}{2\pi x}\).
Direction: curl the right-hand fingers around the wire with the thumb along \(+y\). At a point on the \(+x\) axis, the field points along \(-\hat{k}\).
\[ \vec{B} = -\frac{\mu_0 I}{2\pi x}\,\hat{k} \]

Step 3: Part (a): Force on the charge:
\[ \vec{F} = Q\,(v_0\hat{i}) \times \left(-\frac{\mu_0 I}{2\pi x}\hat{k}\right) = -\frac{\mu_0 I Q v_0}{2\pi x}\,(\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\):
\[ \vec{F} = \frac{\mu_0 I Q v_0}{2\pi x}\,\hat{j} \]
The magnitude is \(\dfrac{\mu_0 I Q v_0}{2\pi x}\), and the force acts along \(+y\), parallel to the current.

Step 4: Part (b): Galvanometer to voltmeter:
A galvanometer can only carry \(I_g = 5\) mA. To measure up to 1 V, connect a high resistance R in series with it. The galvanometer then carries \(I_g\) at 1 V, and the rest of the voltage drops across R.
\[ V = I_g (G + R) \Rightarrow R = \frac{V}{I_g} - G \]
\[ R = \frac{1}{5 \times 10^{-3}} - 100 = 200 - 100 = 100\ \Omega \]

Final Answer:
The force is \(\mu_0 I Q v_0/(2\pi x)\) along \(+y\). A 100 \(\Omega\) resistor in series converts the galvanometer into a 0-1 V voltmeter. \[ \boxed{F=\frac{\mu_0 I Q v_0}{2\pi x}\ (+\hat{j}),\quad R=100\ \Omega} \]