Magnetism – JEE First

Ampere’s law and symmetry

October 31st, 2022jeefirst0 Comments

Problem 3.231 from Irodov.

A current $I$ flows along a lengthy straight wire, as shown in the figure below. From the point O the current spreads radially all over an infinite conducting plane perpendicular to the wire. Find the magnetic field above and below the plane.

Rendered by QuickLaTeX.com — A wire carrying current to an infinite conducting plane. We indicate the radial current on the plane with a few lines, but there is current flowing along every point of the plane.

Solution:

We will use this problem to demonstrate the use of Ampere’s law, starting with some simple scenarios. Our goal is to understand …

Biot-Savart law on a polygon

September 30th, 2022jeefirst0 Comments

A current $I$ flows along a thin wire, shaped as a regular polygon with $N$ sides, which can be inscribed intro a circle of radius $R$ . Find the magnetic field at the center of the polygon. What happens if $N$ is made very large?

Solution:

The polygon described in the problem is illustrated in the figure below, for $N=8$ . The magnetic field created by this structure is the superposition of fields from $N$ straight current carrying wires of length $2 R \sin (\pi/N)$ . Therefore, we will need to find the magnetic field due to a wire of finite length first.

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The Biot-Savart law gives the magnetic field …

Terminal velocity in a magnetic field

February 23rd, 2022jeefirst0 Comments

A copper connector of mass $m$ slides down two smooth copper bars, set at an angle $\alpha$ to the horizontal, due to gravity (see figure). At the top the bars are interconnected through a resistance $R$ . The separation between the bars is $\ell$ . The system is located in a uniform magnetic field of induction $B$ , perpendicular to the plane in which the connector slides. The resistance of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop are assumed to be negilible. If the connector is released from rest at $t=0$ ,

Find the velocty $v(t)$ of

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Electromagnetic induction in a twisted loop

February 21st, 2022jeefirst0 Comments

JEE Advanced 2017 Paper 1, Question 5

A circular insulated copper wire loop is twisted to form two loops of area $A$ and $2 A$ as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field ${\bf B}$ points into the plane of the paper. At $t=0$ , the loop starts rotating about the common diameter as axis with a constant angular velocity $\omega$ in the magnetic field. Which of the following options is/are correct?

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The emf induced in the loop is proportional to the sum

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Conducting wire in a magnetic field

December 30th, 2021jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 6

A conducting wire of parabolic shape, initially $y = x^2$ , is moving with velocity $\vec v = v_0 \hat i$ in a non-uniform magnetic field $\vec B = B_0 \left( 1 + \left( \frac{y}{L} \right)^\beta \right) \hat k$ , as shown in the figure below. If $v_0, B_0, L$ and $\beta$ are positive constants and $\Delta \phi$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

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$|\Delta \phi| = \frac{1}{2} B_0 v_0 L$ for $\beta = 0$ .
$|\Delta \phi| = \frac{4}{3} B_0 v_0 L$ for $\beta = 2$ .
$|\Delta \phi|$ remains the same if the parabolic wire is replaced by a straight wire, $y=x$ initially, of length $\sqrt{2} L$ .
$|\Delta \phi|$ is proportional to the length of the wire projected on the $y$ axis.