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

`T\approx 1.95* 10^(-4) N.m`

Step-by-step 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* 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*10^(-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* 0.45* 1.2* 10^(-3)* 4\pi*10^(-7) * 1400* 4.9* sin 90^(\circ)`

`T\approx 1.95* 10^(-4) N.m`

Answer 2

The torque on the coil is
`1.955* 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* 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* 4\pi times 10^(- 7)* 4.9 = 8.62* 10^(- 3)\ T`

Now, the torque exerted is given by:

`\tau = I'NAB`

`\tau = 0.45* 42* 1.2* 10^(- 3)* 8.62* 10^(- 3) = 1.955* 10^(- 4)\ N-m`