AP EAPCET 07-Jul-2022 Shift 2 Paper

Torque on a magnet (dipole moment m ) when it is placed in region of magnetic field (Intensity B) is τ=M×B
Where, M=M.

^

u

M= magnitude of dipole moment and

^

u

= unit vector in direction of M Now in case I given,
B=˙B

^

i

‌ and ‌M=

M

2

(√3

^

i

+

^

j

)
Hence, torque is τ1=

M

2

(√3

^

i

+

^

j

)×B

^

i

=

MB

2

(

^

j

×

^

i

)=

MB

2

(−

^

k

)
Now, given, |τ1|=0.06N−m
So, 0.06=MB∕2 or MB=0.12 units
Now, when M=M(

^ i +√3 ^ j

2

)
and B=2B

^

j

then, torque will be τ2=M×B
=

M

2

(

^

i

+√3

^

j

)×2B

^

j

=MB(

^

i

×

^

j

)=MB(

^

k

)
∴ Magnitude of torque is |τ2|=MB=0.12N−m