MP Board · Class 12 · Physics · Moving Charges and MagnetismA circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of $0.10\text{ T}$ normal to the plane of the coil. If the current in the coil is $5.0\text{ A}$, what is the: (a) total torque on the coil? (b) total force on the coil? (c) average force on each electron in the coil, given the cross-sectional area of the wire is $10^{-5}\text{ m}^2$ and the free electron density is $10^{29}\text{ m}^{-3}$?

Given Data:

• Number of turns ($N$) = $20$
• Radius of the coil ($R$) = $10\text{ cm} = 0.1\text{ m}$
• Magnetic field ($B$) = $0.10\text{ T}$
• Current ($I$) = $5.0\text{ A}$
• Cross-sectional area ($A$) = $10^{-5}\text{ m}^2$
• Electron density ($n$) = $10^{29}\text{ m}^{-3}$
• Elementary charge ($e$) = $1.6 \times 10^{-19}\text{ C}$

(a) Total torque on the coil

1. Formula: The torque $\tau$ experienced by a current-carrying coil in a uniform magnetic field is given by: $$\tau = N I A B \sin\theta$$ where $A$ is the area of the circular coil ($\pi R^2$) and $\theta$ is the angle between the magnetic field and the normal to the coil's plane.
2. Calculation of Area ($A$): $$A = \pi R^2 = 3.1416 \times (0.1)^2 = 3.1416 \times 0.01 = 0.0314\text{ m}^2$$
3. Calculation of Angle ($\theta$): The magnetic field is normal to the plane of the coil, which means the normal vector of the coil is parallel to the magnetic field. Hence, $\theta = 0^\circ$.
4. Substitution: $$\tau = 20 \times 5.0 \times 0.0314 \times 0.10 \times \sin(0^\circ)$$ Since $\sin(0^\circ) = 0$: $$\tau = 0\text{ N}\cdot\text{m}$$ Answer (a): The total torque on the coil is $0\text{ N}\cdot\text{m}$.

(b) Total force on the coil

1. Concept: A closed current loop placed in a uniform magnetic field experiences a net magnetic force of zero because the forces on opposite segments of the loop cancel out vectorially.
2. Answer (b): The total force on the coil is $0\text{ N}$.

(c) Average force on each electron in the coil

1. Formula for Drift Velocity ($v_d$): The relationship between electric current and drift velocity is: $$I = n e A v_d$$ Rearranging for $v_d$: $$v_d = \frac{I}{n e A}$$
2. Substitution for $v_d$: $$v_d = \frac{5.0}{10^{29} \times (1.6 \times 10^{-19}) \times 10^{-5}}$$ $$v_d = \frac{5.0}{10^{29} - 19 - 5} = \frac{5.0}{10^{5}} = 5.0 \times 10^{-5}\text{ m/s}$$
3. Formula for Magnetic Force on an Electron ($F_e$): The magnetic force on a moving charge in a magnetic field is given by Lorentz force law (magnetic component): $$F_e = q v_d B \sin\phi$$ Since the drift velocity of electrons is perpendicular to the magnetic field, $\phi = 90^\circ$ and $\sin(90^\circ) = 1$: $$F_e = e v_d B$$
4. Substitution for $F_e$: $$F_e = (1.6 \times 10^{-19}\text{ C}) \times (5.0 \times 10^{-5}\text{ m/s}) \times (0.10\text{ T})$$ $$F_e = 1.6 \times 5.0 \times 0.10 \times 10^{-24}$$ $$F_e = 8.0 \times 10^{-25}\text{ N}$$ Answer (c): The average force on each electron is $8.0 \times 10^{-25}\text{ N}$.