Question

A wire of length 26.0 cm carrying a current of 6.50 mA is to be formed into a circular coil and placed in a uniform magnetic field B of magnitude 7.21 mT. If the torque on the coil from the field is maximized, what is the magnitude of that maximum torque

Answer 1

Answer:

The magnitude of the maximum torque is  2.52 x 10⁻⁷ N.M

Explanation:

Given;

length of wire, L = 26.0 cm

current, I = 6.5 mA

magnetic field strength, B = 7.21 mT

Maximum torque on the wire is calculated as;

τ = μ x Β

Where;

μ is dipole moment = NIA

N is number of turns = 1 ( for maximum torque on the coil)

I is current

A is the area of the wire;

2πr = L

r = L / 2π

r = (0.26) / 2π = 0.04138 m

A = πr² = π (0.04138 )² = 0.00538 m²

τ = μ x Β

  = (NIA) x Β

  = ( 1 x 6.5 x 10⁻³ x 0.00538) x 7.21 x 10⁻³

  = 2.52 x 10⁻⁷ N.M

Therefore,  the magnitude of the maximum torque is  2.52 x 10⁻⁷ N.M