If $I_{1}$ and $I_{2}$ are the magnitudes if inclined current in the cases I and II respectively, then.

I_{1} = I_{2}

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I_{1} > I_{2}

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I_{1} < I_{2}

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Nothing can be said

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Solution

The correct option is C $I_{1} < I_{2}$
From the Bio-Savart's law,
$B = \frac{μ_{0}}{4 π} \times \frac{I d l sin θ}{r^{2}}$
$B α I$
In this formula, it is clearly that the magnetic field is directly proportional to the current through the wire.
Since the magnetic field of case $1 = B_{1}$ and the magnetic field of case $2 = B_{2}$
$B_{1} < B_{2}$ (because of the decreasing value of magnetic field)
hence, $I_{1} < I_{2}$

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Similar questions

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

I_{3} = π / 2 \int 0 cos x d x,

I_{1}

I_{2}

x

I_{1} = 1 \int 0 e^{- x} {cos}^{2} x d x

I_{2} = 1 \int 0 e^{- x^{2}} {cos}^{2} x d x

I_{3} = 1 \int 0 e^{- x^{3}} d x

If $I_{1} = \int_{0}^{\frac{π}{2}} x s i n x d x & I_{2} = \int_{0}^{\frac{π}{2}} x c o s x d x$ , then which of the following is true?

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