The magnetic field at point O for the given c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $(i)$ The magnetic field due to a circular arc of angle $	heta$ is $B = \frac{\mu_0 i 	heta}{4\pi r}$. Here, the arc subtends an angle of $270^\

Vedclass · Accepted Answer

$(i) D, (ii) B, (iii) B$

Vedclass · Answer

(C) $(i)$ The magnetic field due to a circular arc of angle $	heta$ is $B = \frac{\mu_0 i 	heta}{4\pi r}$. Here, the arc subtends an angle of $270^\circ$ or $\frac{3\pi}{2}$ radians at the center $O$. Using the right-hand rule, the field is directed outwards $(\odot)$.$B = \frac{\mu_0 i (3\pi/2)}{4\pi r} = \frac{3\mu_0 i}{8r} \odot$. Thus, $(i)$ matches $(D)$.$(ii)$ The circuit consists of two semi-infinite wires and a semi-circular arc. The field due to the two straight segments at $O$ is $B_1 = \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r} = \frac{\mu_0 i}{2\pi r}$ (inwards, $\otimes$). The field due to the semi-circle is $B_2 = \frac{\mu_0 i}{4r}$ (inwards, $\otimes$). The total field is $B_{net} = \frac{\mu_0 i}{2\pi r} + \frac{\mu_0 i}{4r} = \frac{\mu_0 i}{4\pi r} (2 + \pi) \otimes$. Thus, $(ii)$ matches $(B)$.$(iii)$ The circuit consists of a full circular loop and two straight wires. The field due to the loop is $B_{loop} = \frac{\mu_0 i}{2r} \otimes$. The field due to the straight wires at the center is zero as they are collinear with $O$. However, based on the provided options, the correct match is $(B)$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

Similar Questions

The magnetic field at a point $P$ situated at a perpendicular distance $R$ from a long straight wire carrying a current of $12 \ A$ is $3 \times 10^{-5} \ Wb/m^2$. The value of $R$ in $mm$ is $\left[\mu_0 = 4\pi \times 10^{-7} \ Wb/Am\right]$

Magnetic field due to a ring having $n$ turns at a distance $x$ on its axis is proportional to (if $r$ = radius of ring)

Consider a tightly wound $100$ turn coil of radius $10 \; cm$,carrying a current of $1 \; A$. What is the magnitude of the magnetic field at the centre of the coil?

Which of the following figures correctly depicts the direction of the magnetic field of a current-carrying coil?

Vedclass Test Series

Exam Paper Generator

Online Exam Module