Question

A long solenoid has 1400 turns per meter of length, and it carries a current of 4.9 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of 90.0˚ with respect to the axis of the solenoid. The coil consists of 42 turns, has an area of 1.2 × 10-3 m2, and carries a current of 0.45 A. Find the torque exerted on the coil.

Answer 1

Answer:

The torque on the coil is $1.955\times 10^{- 4}\ N-m$

Solution:

No. of turns per meter length, n = 1400 turns\m

Current, I = 4.9 A

Angle, $\theta = 90.0^{\circ}$

No. of turns of coil, N = 42 turns

Area, A = $1.2\times 10^{- 3}m^{2}$

Current in the coil, I' = 0.45 A

Now,

To calculate the exerted torque on the coil:

The magnetic field, B produced inside the coil is given by:

$B = n\mu_{o}I$

$B = 1400\times 4\pi times 10^{- 7}\times 4.9 = 8.62\times 10^{- 3}\ T$

Now, the torque exerted is given by:

$\tau = I'NAB$

$\tau = 0.45\times 42\times 1.2\times 10^{- 3}\times 8.62\times 10^{- 3} = 1.955\times 10^{- 4}\ N-m$

Answer 2

Answer:

$T\approx 1.95\times 10^{-4} N.m$

Explanation:

Given:

A long solenoid having

no. of turns per meter, n =1400

current, I = 4.9 A

A small coil of wire placed inside the solenoid

angle of orientation with respect to the axis of the solenoid, $\theta=90\degree$°

no. of turns in the coil, N = 42

area of the coil, $a= 1.2\times 10^{-3} m^2$

current in the coil, $i =0.45 A$

We have for torque:

$T=n.i.a.B. sin\theta$.......................(1)

∵$B=\mu_{0} .n.I$................................(2)

where:

B= magnetic field

$\mu_0=$The permeability of free space =$4\pi\times10^{-7} T.m.A^{-1}$

Substitute B from eq. (2) into eq. (1) we have:

$T=n.i.a.(\mu_0.N.I ).sin\theta$

putting the respective values in above eq.

$T=42\times 0.45\times 1.2\times 10^{-3}\times 4\pi\times10^{-7} \times 1400\times 4.9\times sin 90^{\circ}$

$T\approx 1.95\times 10^{-4} N.m$