A copper ring of radius R = 8 c m and circular cross-sectional A = 2 m m 2 is in a homogeneous magnetic field whose induction is perpendicular to its plane and changes uniformly. At t = 0 the induction is B 0 = 0 and in t = 0.2 s  it increases to B = 2 T . The angular velocity ω at which the ring should be rotated uniformly in order not to have tensile stress in it at time instant t 1 = 0.1 s is approximately: (mass of the ring m = 9 gram, Resistance = 5 m Ω , self-induction and gravity can be neglected)

Option 1 -

1.2 * 10 3 r a d / s e c

Option 2 -

2.4 * 10 3 r a d / s e c

Option 3 -

4.8 * 10 2 r a d / s e c

Option 4 -

1.7 * 10 2 r a d / s e c ext (1/2)

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago
    Correct Option - 4


    Detailed Solution:

    R = 8 c m , A = 2 m m 2 , B 0 = 0  at   t = 0 s e c

    B = 2  Tat  t = 0.2 s e c

    | ε | = 2 - 0 0.2 π × 64 × 10 - 4  volts

    i = | ε | r = 64 π × 10 - 3 5 × 10 - 3  ampere

    i = 64 π 5  ampere

    d F = i d l B =  magnetic force

    d m ω 2 R = centrifugal force

    i d l B = d m ω 2 R i d l B = m 2 π R d l ω 2 R ω = 2 π i B m = 2 π × 64 π × 1 5 × 9 × 10 - 3 ω = 16 π 3 × 10 r a d / s e c = 170 r a d / s e c

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