Exercise 4.1
A circular coil of wire consisting of $\displaystyle 100$ turns, each of radius $\displaystyle 8.0$ cm carries a current of $\displaystyle 0.40$ A. What is the magnitude of the magnetic field B at the centre of the coil?
Disagrees with the book
This working does not reach the answer printed in NCERT. One of the two is wrong and it has not yet been settled which — check it before you rely on it.
NCERT’s answer
π × \(\displaystyle 10^{-4}\) T ≃ $\displaystyle 3.1$ × \(\displaystyle 10^{-4}\) T
By Ampere's law, the field at the centre of a circular coil of \(\displaystyle N\) turns is
\[B=\frac{\mu_0 N I}{2r} \]
Here \(\displaystyle N=100\), \(\displaystyle I=0.40\ \mathrm{A}\), \(\displaystyle r=8.0\ \mathrm{cm}=0.08\ \mathrm{m}\), \(\displaystyle \mu_0=4\pi\times10^{-7}\ \mathrm{T\,m/A}\).
\[B=\frac{(4\pi\times10^{-7})(100)(0.40)}{2(0.08)}=\pi\times10^{-4}\ \mathrm{T} \]
B ≈ $\displaystyle 3.14$ × $\displaystyle 10$⁻⁴ T, directed along the axis of the coil.
Magnitude (long straight wire):
\[B=\frac{\mu_0 I}{2\pi r}=\frac{(2\times10^{-7})(50)}{2.5} \]
B = $\displaystyle 4$ × $\displaystyle 10$⁻⁶ T.Direction — apply the right-hand thumb rule with the thumb pointing along the current (north → south). Taking East = \(\displaystyle \hat x\), North = \(\displaystyle \hat y\), Up = \(\displaystyle \hat z\) (so \(\displaystyle \hat x\times\hat y=\hat z\)), the current direction is \(\displaystyle -\hat y\) and the point (east of the wire) lies along \(\displaystyle +\hat x\). The field direction is
\[\hat B \propto (-\hat y)\times \hat x = \hat z \]
so the field points vertically upward at the point $\displaystyle 2.5$ m east of the wire.
Force per unit length between two long parallel currents:
\[\frac{F}{L}=\frac{\mu_0 I_A I_B}{2\pi d} \]
With \(\displaystyle I_A=8.0\ \mathrm{A}\), \(\displaystyle I_B=5.0\ \mathrm{A}\), \(\displaystyle d=4.0\ \mathrm{cm}=0.04\ \mathrm{m}\):
\[\frac{F}{L}=\frac{(2\times10^{-7})(8.0)(5.0)}{0.04}=2\times10^{-4}\ \mathrm{N/m} \]
For a $\displaystyle 10$ cm = $\displaystyle 0.10$ m section of wire A:
\[F=(2\times10^{-4})(0.10) \]
F = $\displaystyle 2$ × $\displaystyle 10$⁻⁵ N, attractive (directed toward wire B), since the currents are in the same direction.
The magnetic force \(\displaystyle q\vec v\times\vec B\) is always perpendicular to \(\displaystyle \vec v\); since \(\displaystyle v\) and \(\displaystyle B\) are both constant in magnitude (uniform field, electron enters normal to \(\displaystyle B\)), this force has constant magnitude and always points toward a fixed centre — i.e. it acts purely as a centripetal force, so the electron moves in a circle.Balancing magnetic force and centripetal requirement:
\[evB=\frac{mv^2}{r}\ \Rightarrow\ r=\frac{mv}{eB} \]
With \(\displaystyle e=1.6\times10^{-19}\ \mathrm{C}\), \(\displaystyle m_e=9.1\times10^{-31}\ \mathrm{kg}\), \(\displaystyle v=4.8\times10^{6}\ \mathrm{m/s}\), \(\displaystyle B=6.5\ \mathrm{G}=6.5\times10^{-4}\ \mathrm{T}\):
\[r=\frac{(9.1\times10^{-31})(4.8\times10^{6})}{(1.6\times10^{-19})(6.5\times10^{-4})}=\frac{4.368\times10^{-24}}{1.04\times10^{-22}} \]
r ≈ $\displaystyle 4.2$ × $\displaystyle 10$⁻² m (≈ $\displaystyle 4.2$ cm).A misprint in the printed question. NCERT's question sheet gives the electron's charge as \(\displaystyle 1.5\times10^{-19}\ \mathrm{C}\). The elementary charge is \(\displaystyle 1.6\times10^{-19}\ \mathrm{C}\), and NCERT's own printed answer of $\displaystyle 4.2$ cm uses $\displaystyle 1.6$ — so $\displaystyle 1.6$ is used above. Carrying the misprinted $\displaystyle 1.5$ through instead gives $\displaystyle 4.48$ cm.