Respuesta :
Answer:
a)[tex]T = 2.9*10^{-5} N-m[/tex]
b) north edge will rise up
Explanation:
torque on the coil is given as
[tex]T = NIABsin\theta[/tex]
where N is number of loop = 9 loops
i is current = 7.80 A
-B -earth magnetic field = [tex]5.00*10^{-5} T[/tex]
A- area of circular coil
[tex]A = \frac{\pi d^{2}}{4}[/tex]
[tex]A =\frac{\pi .14^{2}}{4}[/tex]
A =0.015 m2
PUTITNG ALL VALUE TO GET TORQUE
[tex]T = 9*7.8*0.015*5*10^{-5} sin{90-56}[/tex]
[tex]T = 2.9*10^{-5} N-m[/tex]
b) north edge will rise up
Answer:
The torque on the coil is [tex]2.85\times10^{-5}\ N.m[/tex] and North edge is rise up.
Explanation:
Given that,
Number of loops N =9
Diameter d= 14.0 cm
Magnetic field [tex]B= 5.00\times10^{-5}\ T[/tex]
Angle = 56.0°
Current I= 7.80 A
We need to calculate the area of the coil
Using formula of area
[tex]A= \pi r^2[/tex]
[tex]A=\pi\times(7.0)^2[/tex]
[tex]A=0.0154\ m^2[/tex]
(a). We need to calculate the torque on the coil
Using formula of torque
[tex]\tau=NIAB\sin\theta[/tex]
[tex]\tau=9\times7.80\times0.0154\times10^{-4}\times5.00\times10^{-5}\times\sin(90-56)[/tex]
[tex]\tau=9\times7.80\times0.0154\times5.00\times10^{-5}\times\sin34^{\circ}[/tex]
[tex]\tau=2.85\times10^{-5}\ N.m[/tex]
(b). North edge is rise up.
Hence, The torque on the coil is [tex]2.85\times10^{-5}\ N.m[/tex] and North edge is rise up.
