A magnetic needle suspended parallel to a magnetic field requires √3J of work

A Magnetic Needle Suspended Parallel To A Magnetic Field Requires Physics QuestionA Magnetic Needle Suspended Parallel To A Magnetic Field Requires Physics Question

A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be:

Options

(a) 2 √3 J
(b) 3 J
(c) √3 J
(d) 3/2 J

Correct Answer:

3 J

Explanation:

According to work energy theorem
W = U final – U initial = MB (cos 0 – cos 60⁰)
W = MB/2 = √3J …(i)
? = M⃗ x B⃗ = MB sin 60⁰ = (MB √3 / 2) …(ii)
From eq (i) and (ii)
? = 2 √3 x √3 / 2 = 3 J

AddThis Website Tools
admin:

Related Questions

  1. A galvanometer having internal resistance 10Ω requires 0.01 A for a full scale
  2. A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide
  3. Copper of fixed volume V is drawn into wire of length l. When this wire is subjected
  4. A person wants a real image of his own, 3 times enlarged. Where should he stand
  5. A man goes 10 m towards north, then 20m towards east, the displacement is