Question:
A wire carrying current \(I\) is tied between points P and Q and is in the shape of a circular arch of radius \(R\) due to a uniform magnetic field \(B\) (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle \(2\theta_0\) at the centre of the circle (of which it forms an arch) then the tension in the wire is :
A wire carrying current \(I\) is tied between points P and Q and is in the shape of a circular arch of radius \(R\) due to a uniform magnetic field \(B\) (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle \(2\theta_0\) at the centre of the circle (of which it forms an arch) then the tension in the wire is :
Show Hint
The magnetic force on the arc equals that on its chord, \(IB(2R\sin\theta_0)\); balance it against the perpendicular components of the two end tensions, \(2T\sin\theta_0\).
Updated On: Nov 11, 2026
- \(IBR\)
- \(\frac{IBR}{\sin\theta_0}\)
- \(\frac{IBR}{2\sin\theta_0}\)
- \(\frac{IBR\theta_0}{\sin\theta_0}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
The chord PQ has length \(2R\sin\theta_0\). The net magnetic force on a current-carrying curved wire in a uniform field equals that on the straight chord joining its ends: \(F = I B (2R\sin\theta_0)\), perpendicular to the chord (pointing away from the centre, or towards it, depending on direction).
At each end the tension T acts along the tangent, which makes angle \(\theta_0\) with the chord. The components along the chord cancel; the components perpendicular to the chord add up to \(2T\sin\theta_0\), directed against the magnetic force.
Equilibrium: \(2T\sin\theta_0 = 2IBR\sin\theta_0\).
\(T = IBR\).
Answer: (1) \(IBR\)
At each end the tension T acts along the tangent, which makes angle \(\theta_0\) with the chord. The components along the chord cancel; the components perpendicular to the chord add up to \(2T\sin\theta_0\), directed against the magnetic force.
Equilibrium: \(2T\sin\theta_0 = 2IBR\sin\theta_0\).
\(T = IBR\).
Answer: (1) \(IBR\)
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Moving charges and magnetism Questions
- The magnetic field existing in a region is given by \[\vec{B} = 0.2(1 + 2x)\hat{k} \, \text{T}.\]A square loop of edge \( 50 \, \text{cm} \) carrying \( 0.5 \, \text{A} \) current is placed in the \( x \)-\( y \) plane with its edges parallel to the \( x \)- and \( y \)-axes, as shown in the figure. The magnitude of the net magnetic force experienced by the loop is ______ \( \text{mN} \).
- Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is\( \_\_\_\_\_\) cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.)
- An ideal solenoid is kept with its axis vertical. Current $I_0$ is flowing in the solenoid. A charge $Q$ is thrown downward inside the solenoid. The acceleration of the charged particle is then
- For a circular coil of radius \(R\), magnetic field at the center is \(B_0 = 16~\mu\text{T}\). What will be the magnetic field on the axis at a distance \(x = \sqrt{3}R\) from the center?
- The electric current in a circular coil of 2 turns produces a magnetic induction $B _1$ at its centre The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction $B _2$ at its centre The ratio of $\frac{ B _2}{ B _1}$ is :
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions