A long straight wire carrying a current of 30 | vedclass

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

Question

(C) Given: Current $i = 30 \ A$, External magnetic field $B_1 = 4 	imes 10^{-4} \ T$, Distance $r = 2 \ cm = 2 	imes 10^{-2} \ m$.The magnetic field

Vedclass · Accepted Answer

$5 	imes 10^{-4}$

Vedclass · Answer

(C) Given: Current $i = 30 \ A$, External magnetic field $B_1 = 4 	imes 10^{-4} \ T$, Distance $r = 2 \ cm = 2 	imes 10^{-2} \ m$.The magnetic field produced by the straight wire at a distance $r$ is given by $B_2 = \frac{\mu_0 i}{2 \pi r}$.Substituting the values: $B_2 = \frac{2 	imes 10^{-7} 	imes 30}{2 	imes 10^{-2}} = 3 	imes 10^{-4} \ T$.Since the external magnetic field $B_1$ is parallel to the wire, the magnetic field $B_2$ produced by the wire (which is tangential to the circular path around the wire) is perpendicular to $B_1$.Therefore, the resultant magnetic field $B$ is $B = \sqrt{B_1^2 + B_2^2}$.$B = \sqrt{(4 	imes 10^{-4})^2 + (3 	imes 10^{-4})^2} = \sqrt{16 	imes 10^{-8} + 9 	imes 10^{-8}} = \sqrt{25 	imes 10^{-8}} = 5 	imes 10^{-4} \ T$.

Similar Questions

Which of the following graphs correctly represents the variation of magnetic field $B$ with distance $R$ from a long straight current-carrying conductor?

Identical currents flow in two perpendicular wires,as shown in the figure. The wires are very close but do not touch. The magnetic field can be zero:

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?

$A$ straight conductor carries a current of $5 \, A$. An electron travelling with a speed of $5 \times 10^6 \, m/s$ parallel to the wire at a distance of $0.1 \, m$ from the conductor,experiences a force of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module