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

Answer:

The torque on the coil is $2.85\times10^{-5}\ N.m$ and North edge is rise up.

Explanation:

Given that,

Number of loops N =9

Diameter d= 14.0 cm

Magnetic field $B= 5.00\times10^{-5}\ T$

Angle = 56.0°

Current I= 7.80 A

We need to calculate the area of the coil

Using formula of area

$A= \pi r^2$

$A=\pi\times(7.0)^2$

$A=0.0154\ m^2$

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

Using formula of torque

$\tau=NIAB\sin\theta$

$\tau=9\times7.80\times0.0154\times10^{-4}\times5.00\times10^{-5}\times\sin(90-56)$

$\tau=9\times7.80\times0.0154\times5.00\times10^{-5}\times\sin34^{\circ}$

$\tau=2.85\times10^{-5}\ N.m$

(b). North edge is rise up.

Hence, The torque on the coil is $2.85\times10^{-5}\ N.m$ and North edge is rise up.

Answer 2

Answer:

a)$T = 2.9*10^{-5} N-m$

b) north edge will rise up

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

$A = \frac{\pi d^{2}}{4}$

$A =\frac{\pi .14^{2}}{4}$

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