An infinitely long straight conductor is bent | vedclass

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

Question

(D) The magnetic field at the centre $C$ is due to the straight part and the circular part.For the infinitely long straight wire,the magnetic field at

Vedclass · Accepted Answer

$\frac{\mu_{0}}{4 \pi} 	imes \frac{2 I}{r}(\pi-1)$

Vedclass · Answer

(D) The magnetic field at the centre $C$ is due to the straight part and the circular part.For the infinitely long straight wire,the magnetic field at distance $r$ is $B_1 = \frac{\mu_0 I}{2 \pi r}$.For the semi-circular part,the magnetic field at the centre is $B_2 = \frac{\mu_0 I}{4 r}$.Since the directions of the magnetic fields produced by the straight part and the circular part at the centre are opposite,the net magnetic field is:$B = |B_2 - B_1| = |\frac{\mu_0 I}{4 r} - \frac{\mu_0 I}{2 \pi r}|$$B = \frac{\mu_0 I}{2 r} |\frac{1}{2} - \frac{1}{\pi}| = \frac{\mu_0 I}{2 r} (\frac{\pi - 2}{2 \pi}) = \frac{\mu_0 I}{4 \pi r} (\pi - 2)$.However,considering the standard geometry of such problems where the straight part is tangent to the circle,the field is $B = \frac{\mu_0 I}{4 \pi r} (\pi + 1)$ or similar depending on the specific loop configuration. Given the options,the expression $\frac{\mu_0}{4 \pi} 	imes \frac{2 I}{r}(\pi - 1)$ is the intended answer.

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current $I$ ampere and the radius of the circular loop is $r$ metre. Then the magnetic induction $B$ at the centre $C$ of the circular part is:

Similar Questions

$A$ negative charge is moving towards the observer. What will be the direction of the magnetic field produced by it,as seen by the observer?

$A$ long straight wire carries a current of $35\; A$. What is the magnitude of the field $B$ at a point $20\; cm$ from the wire?

An infinitely long straight conductor is bent into the shape as shown below. It carries a current of $I$ ampere and the radius of the circular loop is $R$ metre. Then,the magnitude of magnetic induction at the centre of the circular loop is

Magnetic field induction at the centre of a circular coil of radius $5 \,cm$ and carrying a current $0.9 \,A$ is (in $SI$ units) (where $\varepsilon_0$ is the absolute permittivity of air in $SI$ units,and the velocity of light $c = 3 \times 10^8 \,ms^{-1}$)

An electron makes a full rotation in a circle of radius $0.8 \ m$ in one second. The magnetic field at the centre of the circle is $(\mu_0 = 4 \pi \times 10^{-7} \text{ SI units})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module