Question

A single circular loop of radius 0.22m carries a current of 2.6A in a magnetic field of 0.95T .

Part A

What is the maximum torque exerted on this loop?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A single circular loop of radius 0.22m carries a current of 2.5A in a magnetic field of 0.94T.

What is the maximum torque exerted on this loop?

Answer

the torque on a current carrying wire in a magnetic field is:

T= N I A B sin?

where N=number of loops (here N=1)
I = current, here 2.5 amp
A is the area of the loop=? r^2 where r=0.22m
B=mag field strength, =0.94T
?= angle between field and normal to loop; t

he max torque occurs when ?=90 such that sin ? =1,

T=NIAB =1*2.5*?*0.22²*0.94 =0.357N

Answer 2

1st the force(F)=NBIL where N is number of loops N=1 since there is single loop,B is magnetic field, I is current in the wire and L is the radius
but Torque(T) =F*L
T= BIL*L
T = 0.95*2.6*0.22*0.22
T = 0.119548Nm

Answer 3

$\tau =\mu *B=i*A*B=2.6*\pi *0.22^{2}*0.950=0.375N-m$