An infinite straight current-carrying conduct | vedclass

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Question

(D) The magnetic field at the center of the loop is the sum of the magnetic field due to the straight wire and the magnetic field due to the circular

Vedclass · Accepted Answer

$\frac{\mu_0}{4 \pi} \frac{2 I}{R}(\pi+1)$

Vedclass · Answer

(D) The magnetic field at the center of the loop is the sum of the magnetic field due to the straight wire and the magnetic field due to the circular loop.Magnetic field at the center due to the straight wire at a distance $R$ is given by:$B_{	ext{wire}} = \frac{\mu_0 I}{2 \pi R} = \frac{2 \mu_0 I}{4 \pi R}$ (directed inwards).Magnetic field at the center due to the circular loop of radius $R$ is given by:$B_{	ext{loop}} = \frac{\mu_0 I}{2 R} = \frac{\mu_0 I}{2 R} 	imes \frac{2 \pi}{2 \pi} = \frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I}{R}$ (directed inwards).Since both magnetic fields are in the same direction,the net magnetic field $B$ is:$B = B_{	ext{wire}} + B_{	ext{loop}}$$B = \frac{2 \mu_0 I}{4 \pi R} + \frac{2 \mu_0 I \pi}{4 \pi R}$$B = \frac{\mu_0}{4 \pi} \frac{2 I}{R} (\pi + 1)$ (directed inwards).Thus,the correct option is $D$.

An infinite straight current-carrying conductor is bent in such a way that a circular loop is formed on it as shown in the figure. If the radius of the loop is $R$,the magnetic field at the centre of the loop is . . . . . . .

Similar Questions

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