can anyone answer to this physics question?

Question

A current goes around a loop.
a)Find the torque about the vertical axis.
b)Express the torque in terms of the magnetic moment of the current loop.
c)If B=1.5T, I =10A, and the loop is 3*5 cm, find the torque?

Answer 1

a)
consider a reactangular loop PQRS suspended in a uniform B. let PQ=RS=l and QR=SP=b. let current I flowing through the coil in the dir. PQRS and @ be the angle of the plane of the coil to the dir of the B.

let f1,f2,f3,f4 be the force acting on 4 arms PQ,QR,RS,SP of the coil.
the force on SP is:
f4=I(spXB)
and f4=I(SP)Bsin(180-@)
=IbBsin@
the dir of this force is in the dir of (SPXB) i.e. in the plane of coil dir upwards.
the force on arm QR is:
f2=I(QRXB)
or, f2=I(QR)Bsin@=IbBsin@
the dir of this force is in the plane of the coil dir downwards

since the force f2 and f4 are equal in magnitude and acting in opp dir along the same st. line, they cancel out each other.
now,
the force on PQ is
f1=I(PQXB)
or, f1=I(PQ)Bsin90=IlB
(since PQ is prepandicuar to B)
acc. to fleming,s left hand rule, the dir of this force is prepandicular to the plane of the loop dir outwards
and force on RS is
f3=I(RSXB)
or, f3=I(RS)Bsin90=IlB
(since RS is prepandicular to B)
the dir is prepandicular to the plane of paper away from reader by same rule
the torque on the coil is
t=either force X arm of the couple
=IlB X PScos@
=IBAcos@
(since A is the area of the loop)
or, t=nIBAsin#=MBsin# = I M X B I
as #+@=90 (and n is the no. of turns)


b)
B=1.5 T
I=10A
A=3*5=15
as you don't mention the no. of the turn so i took it as 1
n=1
as you don't mention the angle so i took it as 90
then sin90=1
now,
t=nIBAsin#
=1*10*1.5*15*1=225 N-m (ans)