Question

A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.1 tesla normal to the plane of the coil if the current in the coil is 5 ampere what is the total torque on the coil total force on the coil average force on each electron in the coil due to magnetic field

Answer 1

$\mathfrak{\huge{\underline{\underline{AnswEr :}}}}$

Cross-sectional area of copper coil, A = 10⁼⁵ m²

Number of free electrons per cubic meter in copper, N = 10²⁹ /m³

Charge on the electron, e = 1.6 × 10⁻¹⁹ C

Magnetic force, F = Bevd

Where,

vd = Drift velocity of electrons

= l/NeA

» F = Bel/NeA

= 0.10 × 5.0/10²⁹ × 10⁼⁵

Hence, the average force on each electron is 5 x 10⁼²⁵ N.

Answer 2

$\huge\mathcal{Bonjour}$

⭑A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.1 tesla normal to the plane of the coil if the current in the coil is 5 ampere what is the total torque on the coil total force on the coil average force on each electron in the coil due to magnetic field:-

⭒Number of turns on the circular coil, n = 20

Radius of the coil,r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

(a) The total torque on the coil is zero because the field is uniform.

(b) The total force on the coil is zero because the field is uniform.

(c) Cross-sectional area of copper coil, A = 10-5 m2

Number of free electrons per cubic meter in copper, N = 1029 /m3

Charge on the electron, e = 1.6 × 10-19 C

Magnetic force, F = Bevd

Where,

vd = Drift velocity of electrons

[Now,see the attachment]

Hence, the average force on each electron is 5 x 10^-25 N.

Thanks!!