Question

A circular loop of radius 11.9 cm is placed in a uniform magnetic field.

(a) If the field is directed perpendicular to the plane of the
loop and the magnetic flux through the loop is 7.40 ✕
10^{−3} T · m^{2}, what is the strength of the
magnetic field?

T

(b) If the magnetic field is directed parallel to the plane of the
loop, what is the magnetic flux through the loop?

T · m^{2}

Answer #1

Here it is given that magnetic field is directed perpendicularly to the plane , so flux will be maximum, Since the angle between normal to the loop and magnetic field θ=0

MAGNETIC FLUX,

Φ =BACos θ

Here θ=0 . Therefore

MAGNETIC FLUX,

Φ = BA, where B= magnetic field, A=area

Now we have given value of flux

Φ =7.40×10^{-3}Tm2

Now we have radius r=11.9cm=0.119m

Therefore area A=πr²=π×(0.119)²=0.04446m²

Cos0°=1

Substituting all the values

Φ =BACos0°

7.40×10^{-3}=B×0.04446

B=7.40×10^{-3}/0.04446

B=0.16644=166.44×10^{3}T

b) since magnetic field is parallel to the plane therefore the value of θ=90° which causes the flux to be minimum as zero since cos90=0

Therefore, flux through the loop is zero

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