A coil is placed in a time varying magneitc field. The power dissipated due to current induced in the coil is `P_(1)`. If the number of turns is doubled and radius of the wire is halved, the power dissipated is `P_(2)`. Then `P_(1) : P_(2)` is

Options

A. `1:4`
B. `3:2`
C. `2:1`
D. `4:1`

Answer

A

Explanation

According to the question,
radius of wire become `(r )/(2)`, so its length will be 4l
and its resistance will become 16 R .
The number of turns is doubled in a coil, so its
radius should be doubled to accommodate the length of wire, the area of coil will become 4 times .
`rArr " " V_(2) = 8V_(1)" " ` ... (i)
Now, current induced in the coil is P,
` therefore " " p_(1) = (v_(1)^(2))/(R)`
and power dissipated ,` p_(2) = (V_(2)^(2))/(16R) " " ` ... (ii)
From Eqs. (i) and (ii), we get
`P_(2) = ((8V_(1))^(2))/(16R) rArr P_(2) = (64 V_(1)^(2))/(16R) `
`therefore " " P_(2) = (4V_(1)^(2))/(R )`
Then the ratio, `P_(1) : P_(2) = (V_(1)^(2))/(R ) : (4V_(1)^(2))/(R ) or P_(1) : P_(2) = 1 : 4 `