A long, straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2 a,$ respectively from the axis of the wire is
JEE MAIN 2020, Diffcult
Download our app for free and get started
Let current density be J.
$\therefore$ Applying Ampere's law.
$\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_{0} \mathrm{i} \Rightarrow \mathrm{B}_{\mathrm{A}} 2 \pi \frac{\mathrm{a}}{3}=\mu_{0} \mathrm{J} \pi\left(\frac{\mathrm{a}}{3}\right)^{2}$
$\therefore B_{A}=\frac{\mu_{0} J a}{6}$
Similarly, $B_{B}=\frac{\mu_{0} J a}{4}$
$\frac{B_{A}}{B_{B}}=\frac{\mu_{0} J a \times 4}{\mu_{0} J 6 a}=\frac{2}{3}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A small square loop of wire of side $l$ is placed inside a large square loop of wire of side $L(L >> l)$. The loops are coplaner and their centres coincide. The mutual inductance of the system is propotional toView Solution
- 2The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is ${B_1}$. The field at the centre of another coil of radius $2 r$ carrying same current $I$ is ${B_2}$. The ratio $\frac{{{B_1}}}{{{B_2}}}$ isView Solution
- 3Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire $C$ which carries a current of $5.0\, amp$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is thenView Solution
- 4In an ammeter, $5 \%$ of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be :View Solution
- 5View SolutionIn a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a
- 6Two mutually perpendicular insulated conducting wires carrying equal currents $I$, intersect at origin. Then the resultant magnetic induction at point $P(2m, 3m)$ will beView Solution
- 7A particle is moving with velocity $\overrightarrow{ v }=\hat{ i }+3 \hat{ j }$ and it produces an electric field at a point given by $\overrightarrow{ E }=2 \hat{ k }$. It will produce magnetic field at that point equal to (all quantities are in SI units)View Solution
- 8Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.View Solution
Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$, the voltmeter with resistance $1000\,\Omega$ will be preferred over voltmeter with resistance $4000\,\Omega$.
Reason $R:$ Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below.
- 9View SolutionA galvanometer can be converted into an ammeter by connecting
- 10A triangular loop of side $l$ carries a current $I$. It is placed in a magnetic field $B$ such that the plane of the loop is in the direction of $B$. The torque on the loop isView Solution