Question

A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.00×10−5T and points into the Earth at an angle of 56.0 ∘ below a line pointing due north. A 7.80-A clockwise current passes through the coil.(a) determine the torque on the coil, and(b) which edge of the coil rises up: north, east, south, or west?

Answer 1

The torque on the coil is
`2.85*10^(-5)\ N.m` and North edge is rise up.

Step-by-step explanation:

Given that,

Number of loops N =9

Diameter d= 14.0 cm

Magnetic field
`B= 5.00*10^(-5)\ T`

Angle = 56.0°

Current I= 7.80 A

We need to calculate the area of the coil

Using formula of area

`A= \pi r^2`

`A=\pi*(7.0)^2`

`A=0.0154\ m^2`

(a). We need to calculate the torque on the coil

Using formula of torque

`\tau=NIAB\sin\theta`

`\tau=9*7.80*0.0154*10^(-4)*5.00*10^(-5)*\sin(90-56)`

`\tau=9*7.80*0.0154*5.00*10^(-5)*\sin34^(\circ)`

`\tau=2.85*10^(-5)\ N.m`

(b). North edge is rise up.

Hence, The torque on the coil is
`2.85*10^(-5)\ N.m` and North edge is rise up.

Answer 2

a)
`T = 2.9*10^(-5) N-m`

b) north edge will rise up

Step-by-step explanation:

torque on the coil is given as

`T = NIABsin\theta`

where N is number of loop = 9 loops

i is current = 7.80 A

-B -earth magnetic field =
`5.00*10^(-5) T`

A- area of circular coil

`A = (\pi d^(2))/(4)`

`A =(\pi .14^(2))/(4)`

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

`T = 9*7.8*0.015*5*10^(-5) sin{90-56}`

`T = 2.9*10^(-5) N-m`

b) north edge will rise up