Question

A circular coil with 169 turns has a radius of 2.6 cm. (a) What current through the coil results in a magnetic dipole moment of 3.4 A·m2? I= A (b) What is the maximum torque that the coil will experience in a uniform field of strength 0.03 T? τmax= Nm (c) If the angle between μ and B is 44.9∘, what is the magnitude of the torque on the coil? τ= Nm (d) What is the magnetic potential energy of coil for this orientation?

Answer 1

Answer with Explanation:

We are given that

Number of turns=n=169

Radius of coil=r=2.6 cm=$2.6\times 10^{-2} m$

$1 cm=10^{-2}m$

Area of circular coil=$\pi r^2$

Where $\pi=3.14$

Using the formula

Area of circular coil=$3.14\times (2.6\times 10^{-2})^2=21.2\times 10^{-4}m^2$

a.Magnetic dipole  moment=$\mu=3.4 Am^2$

$I=\frac{\mu}{nA}$

Using the formula

$I=\frac{3.4}{169\times 21.2\times 10^{-4}}$

$I=9.5 A$

b.Magnetic field,  B=0.03 T

$\tau_{max}=\mu B$

Substitute the value

$\tau_{max}=3.4\times \times 0.03=0.102 Nm$

c.$\theta=44.9^{\circ}$

$\tau=\tau_{max}sin\theta$

Substitute the values

$\tau=0.102sin44.9=0.07 Nm$