A long circular tube of length `10 m` and radius ` 0.3 m` carries a current `I` along its curved surface as shown . A wire - loop of resistance `0.005 ohm` and of radius `0.1 m` is placed inside the tube its axis coinciding with the axis of the tube . The current varies as `I = I_(0)cos(300 t)` where `I_(0)` is constant. If the magnetic moment of the loop is `Nmu_(0)I_(0) sin (300 t)`, then 'N' is

Answer

d

Explanation

According to Amper's circuital law the magnetic field inside the tube is `B=(mu_(0))/(L)" "....(i)`
Where L is the length of the tube.
Flix linked with the wire loop is `phi=Bpir^(2)`
Where r is the radius of the loop
`phi(mu_(0))I)/(L)pir^(2)" "` ( Using (i))
`=(mu_(0)pir^(2)L_(0)cos 300t)/(L)`
Induced emf in the loop is `epsi=-(dphi)/(dt)=-(d)/(dt)((mu_(0))/(L)pir^(2)I_(0)cos300t)`
`=(mu_(0)pir^(2)I_(0)300sin 300t)/(L)`
Induced current in the loop is `i=(epsi)/(R)=(300mu_(0)pir^(2)I_(0)sin300t)/(LR)`
where R is th resistance of the loop Mgnetic monent of the loop `M=i pir^(2)`
`=(300pi^(2)r^(4)mu_(0)I_(0)sin300t)/(LR)`
Substituting hte given values, we get
`M=(300xx10xx(0.1)^(4))/(10xx0.005)mu_(0)I_(0)sin300t("Take"pi^(2)=10)`
`=6mu_(0)I_(0)sin300t`
`M=Nmu_(0)I_(0)sin300t`
`:.N=6`