Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu _0 = 4p\times10^{-7}\, N-A^{-2})$
JEE MAIN 2019, Diffcult
Download our app for free and get startedPlay store
$B=\frac{\mu_{0} I}{4 \pi d} 2 \sin \theta$

$d=4\, \mathrm{cm}$

$\sin \theta=\frac{3}{5}$

art

Download our app
and 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

  • 1
    The net charge in a current carrying wire is zero still magnetic field exerts a force on it, because a magnetic field exerts force on
    View Solution
  • 2
    A current carrying wire in the neighborhood produces
    View Solution
  • 3
    Consider the following statements regarding a charged particle in a magnetic field . Which of the statements are true :
    View Solution
  • 4
    The magnetic induction at the centre of a current carrying circular coil of radius $10\, cm$ is $5\sqrt 5 \,times$ the magnetic induction at a point on its axis. The distance of the point from the centre of the coil (in $cm$) is
    View Solution
  • 5
    An ammeter of $100$ $\Omega$ resistance gives full deflection for the current of $10^{-5} \,amp$. Now the shunt resistance required to convert it into ammeter of $1\, amp$. range, will be
    View Solution
  • 6
    In an ionised sodium atom, an electron is moving in a circular path of radius $r$ with angular velocity $\omega $. The magnetic induction in $wb/m^2$ produced at the nucleus will be
    View Solution
  • 7
    A charged particle moves in a magnetic field $\vec B = 10\,\hat i$ with initial velocity $\vec u = 5\hat i + 4\hat j$ The path of the  particle will be
    View Solution
  • 8
    The figure shows a circular loop of radius a with two long parallel wires (numbered $1$ and $2$) all in the plane of the paper. The distance of each wire from the centre of the loop is $d$. The loop and the wires are carrying the same current $I$. The current in the loop is in the counterclockwise direction if seen from above.$Image$

    $1.$ When $d \approx$ a but wires are not touching the loop, it is found that the net magnetic filed on the axis of the loop is zero at a height $h$ above the loop. In that case

    $(A)$ current in wire $1$ and wire $2$ is the direction $P Q$ and $R S$, respectively and $h \approx a$

    $(B)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $SR$, respectively and $h \approx a$

    $(C)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $SR$, respectively and $h \approx 1.2 a$

    $(D)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $RS$, resepectively and $h \approx 1.2 a$

    $2.$ Consider $d \gg a$, and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

    $(A)$ $\frac{\mu_0 I^2 a^2}{d}$ $(B)$ $\frac{\mu_0 I^2 a^2}{2 d}$ $(C)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{d}$ $(D)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{2 d}$

    Give the answer question $1$ and $2.$

    View Solution
  • 9
    An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicularly, Then
    View Solution
  • 10
    A $72 \; \Omega$ galvanometer is shunted by a resistance of $8 \; \Omega$. The percentage of the total current which passes through the galvanometer is $.....$
    View Solution