The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is

Question

The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is (A) (NAB/2) (B) NAB (C) (NAB/4) (D) 0

Tardigrade Admin · Accepted Answer

φ=M B A cos θ =M B A cos 190° φ =-M B A

Q. The magnetic flux linked with a coil of $N$ turns of area of cross section $A$ held with its plane parallel to the field $B$ is

$\frac{N A B}{2}$

$N A B$

$\frac{N A B}{4}$

0

2NAB

$ϕ = MB A cos θ = MB A cos 19 0^{\circ}$
$ϕ = - MB A$

Q. The magnetic flux linked with a coil of N turns of area of cross section A held with its plane parallel to the field B is

2NAB​

NAB

4NAB​

0

2NAB

ϕ=MBAcosθ=MBAcos190∘ ϕ=−MBA

Q. The magnetic flux linked with a coil of $N$ turns of area of cross section $A$ held with its plane parallel to the field $B$ is

$\frac{N A B}{2}$

$N A B$

$\frac{N A B}{4}$

$ϕ = MB A cos θ = MB A cos 19 0^{\circ}$
$ϕ = - MB A$