Question

A wire of length 25.0 cm carrying a current of 4.21 mA is to be formed into a circular coil and placed in a uniform magnetic field B with arrow of magnitude 5.55 mT. Suppose the torque on the coil from the field is maximized.

Answer 1

1.162 x 10^-7 Nm

Step-by-step explanation:

length of wire, l = 25 cm

l = 2 π r

where, r is the radius of circular loop

25 = 2 x 3.14 x r

r = 3.98 cm

Magnetic field, B = 5.55 mT = 5.55 x 10^-3 T

Current, i = 4.121 mA = 4.21 x 10^-3 A

Torque, τ = i x A x B

τ = 4.21 x 10^-3 x 3.14 x 0.0398 x 0.0398 x 5.55 x 10^-3

τ = 1.162 x 10^-7 Nm

Thus, the maximum torque in the coil is 1.162 x 10^-7 Nm.