A coil of 100 turns and area 5 text{ cm}^2 is | vedclass

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Question

(A) The magnetic flux $\phi$ linked with a coil of $N$ turns is given by the formula: $\phi = N B A \cos 	heta$,where $	heta$ is the angle between t

Vedclass · Accepted Answer

$5 	imes 10^{-3} 	ext{ Wb}$

Vedclass · Answer

(A) The magnetic flux $\phi$ linked with a coil of $N$ turns is given by the formula: $\phi = N B A \cos 	heta$,where $	heta$ is the angle between the normal to the coil and the magnetic field direction.Given:$N = 100$$B = 0.2 	ext{ T}$$A = 5 	ext{ cm}^2 = 5 	imes 10^{-4} 	ext{ m}^2$$	heta = 60^o$Substituting the values:$\phi = 100 	imes 0.2 	imes (5 	imes 10^{-4}) 	imes \cos(60^o)$$\phi = 20 	imes 5 	imes 10^{-4} 	imes 0.5$$\phi = 100 	imes 10^{-4} 	imes 0.5$$\phi = 0.5 	imes 10^{-2} 	ext{ Wb} = 5 	imes 10^{-3} 	ext{ Wb}$.

$A$ coil of $100$ turns and area $5 \text{ cm}^2$ is placed in a magnetic field $B = 0.2 \text{ T}$. The normal to the plane of the coil makes an angle of $60^o$ with the direction of the magnetic field. The magnetic flux linked with the coil is:

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