Chapter 20: Q54P (page 859)

In Figure 20.116 a battery with known emf=K is connected to two large parallel metal plates. Each plate has a length L and width W, and the plates are a very short distance apart. The plates are surrounded by a vertical thin circular coil of radius R containing N turns through which runs a steady conventional current I. The center of the coil is at the center of the gap between the plates. At a certain instant, a proton (charge +e, mass M) travels through the center of the coil to the right with speed v, and the net force on the proton at this instant is zero (neglecting the very weak gravitational force). What are the magnitude and direction of conventional current in the coil? Explain clearly.

Short Answer

Expert verified

$I = \frac{2 R K e}{μ_{0} N s v}$ ,clockwise

Step by step solution

Given data

$e m f = K$

Plate has a length $L$ and width $W$

Gap between plates $s$

Coil of radius $R$ containing $N$ turns

Proton charge $+ e$ and mass $M$

Concept/ Formula used

The magnetic field runs parallel to the wire in a perpendicular direction. The direction in which the fingers would curl if you wrapped your right hand's fingers around the wire with your thumb pointing in the direction of the current would indicate the direction of the magnetic field.

Magnetic force ${\vec{F}}_{m} = q \vec{V} \times \vec{B}$

Where $q$ is charge

$\vec{V}$ Is velocity of charge particle and

$\vec{B}$ Is magnetic field

Calculation for conventional current

$\begin{array}{l} F = q (\vec{E} + \vec{v} \times \vec{B}) \\ 0 = q (- E \hat{j} + v \hat{i} \times \vec{B}) \\ v \hat{i} \times \vec{B} = E \hat{j} \\ v \hat{i} \times \vec{B} (- \hat{k}) = E \hat{j} \end{array}$