9303753286127016 A circular coil of n turns is kept in a uniform magnetic field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with the coil is now ϕ . Now the coil is opened and made into another circular coil of twice the radius of the previous coil and kept in the same field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with this coil now is: A ϕ B 2ϕ C 4 ϕ D 2 ϕ

A circular coil of n turns is kept in a uniform magnetic field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with the coil is now ϕ . Now the coil is opened and made into another circular coil of twice the radius of the previous coil and kept in the same field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with this coil now is: A ϕ B 2ϕ C 4 ϕ D 2 ϕ

byBalkishan Agrawal -June 17, 2022

0

Question

A circular coil of n turns is kept in a uniform magnetic field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with the coil is now $ϕ$ . Now the coil is opened and made into another circular coil of twice the radius of the previous coil and kept in the same field such that the plane of the coil is perpendicular to the field. The magnetic flux associated with this coil now is:

A

$ϕ$

B

2 $ϕ$

C

$\frac{ϕ}{4}$

D

$\frac{ϕ}{2}$

Medium

Solution

Verified by GMS

Correct option is B)

$f l u x (ϕ) = n B A$
$= n B π r^{2}$
Let total length of coil is $l$
$l = n (2 π r)$
Now, since same wire length has double radius 80,
$l = n_{1} (2 π 2 r)$
$l = n_{1} = \frac{n}{2}$
Now, flux $= \frac{n}{2} B A$
$= \frac{n}{2} B π 4 r^{2}$
$= 2 n B π r^{2}$
$= 2 ϕ$
So, new flux $= 2 ϕ$

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