A ring of mean radius $15$ cm has $3500$ turns of wire wound on a ferromagnetic core of relative permeability $800$ . The magnetic field in the core for a magnetising current of $1.2$ A is

2.48

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3.48

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4.48

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5.48

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Solution

The correct option is D $4.48$ T
Given,
$r = 15 c m = 0.15 m$
$N = 3500$
$μ_{r} = 800$
$i = 1.2 A$
Magnetic field in the core,
$B = \frac{μ_{0} μ_{r} N i}{2 π r}$
$B = \frac{800 \times 4 π \times 10^{- 7} \times 3500 \times 1.2}{2 π \times 0.15}$
$B = 4.48 T$
The correct option is C.

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A ring of mean radius 15cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. The magnetic field in the core for a magnetising current of 1.2A is

The correct option is D 4.48TGiven,r=15cm=0.15mN=3500μr=800i=1.2AMagnetic field in the core, B=μ0μrNi2πrB=800×4π×10−7×3500×1.22π×0.15B=4.48TThe correct option is C.

A ring of mean radius $15$ cm has $3500$ turns of wire wound on a ferromagnetic core of relative permeability $800$ . The magnetic field in the core for a magnetising current of $1.2$ A is

The correct option is D $4.48$ T
Given,
$r = 15 c m = 0.15 m$
$N = 3500$
$μ_{r} = 800$
$i = 1.2 A$
Magnetic field in the core,
$B = \frac{μ_{0} μ_{r} N i}{2 π r}$
$B = \frac{800 \times 4 π \times 10^{- 7} \times 3500 \times 1.2}{2 π \times 0.15}$
$B = 4.48 T$
The correct option is C.