A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle between its magnetic dipole moment and the field is 30.1∘. Find the strength B of the magnetic field that causes a torque of 1.51×10−5 N⋅m to act on the loop when a current of 2.47 mA flows in it

Question

contestada

Respuesta :

mavila18 mavila18 · Answer 1 · 2020-05-15T03:30:28+02:00

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

[tex]\tau=NIABsin\theta[/tex] (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

[tex]B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T[/tex]