A coil having magnetic moment 15 A m² placed in a uniform magnetic field of 4 T in the horizontal direction exists such that initially the axis of coil is in the direction of the field. If the coil is rotated by `45^(@)` and the moment of inertia of the coil is `0.5" kg m"^(2)` then the angular speed acquired by the coil is

Options

A. `20 "rad s"^(-1)`
B. `10 "rad s"^(-1)`
C. `8.34 "rad s"^(-1)`
D. `4.5 "rad s"^(-1)`

Answer

C

Explanation

Since `tau=I(d omega)/(dt)=mB sin theta" ...(i)"`
Now, `(d omega)/(dt)=(domega)/(d theta)=omega(domega)/(d theta)" ...(ii)"`
then from (i) and (ii)
`Iomega=(domega)/(d theta)=m B sin theta, Iomega d omega=m B sin theta d theta`
Integrating both sides
`Iint_(0)^(omega)omega domega=nBint_(0)^(45^(@))sin theta d theta`
`I[(omega^(2))/(2)]_(0)^(omega_(f))=mB[-cos theta]_(0^(@))^(45^(@))`
`(1)/(2)Iomega_(f)^(2)=-mB[cos45^(@)-cos 0^(@)]`
`=-mB[(1)/(sqrt2)-1]=0.29mB`
`omega_(f)^(2)=(2xx0.29mB)/(0.50)`
`omega_(f)^(2)=69.6, omega_(f)=sqrt(69.6)="8.34 rad s"^(-1)`