A magnetic needle suspended parallel to a magnetic field
requires √3 J of work to turn it through 60°. The torque
needed to maintain the needle in this position will be
(a) 2 √3J (b) 3J (c) √3J (d) 3/2 J
Answers
Answered by
0
Explanation:
W=MB(cosθ
1
−cosθ
2
)
3
=MB(coso
o
−cos60
o
)
3
=
2
MB
.......(1)
τ=MBsinθ
τ=MBsin60
o
=
3
2
MB
.......(2)
∴ τ=(
3
)(
3
)=3J
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2
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