Two long conductors, separated by a distance d carry currents I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. The new value of the force between them is
A wire PQ carries a current ‘i’ is placed perpendicular to a long wire XY carrying a current
I. The direction of force on PQ will be:
A rectangular loop carrying a current i1, is situated near a long straight wire carrying a steady current i2. The wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the figure. Then the current loop will :
A current carrying wire AC is placed in uniform transverse magnetic field then the force on wire AC :
A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (\(g = 9.8\text{ ms}^{-2}\))
For the wire to be suspended, the upward magnetic force must equal gravity: \(I L B = mg ⇒ I = \frac{mg}{LB}\). Plugging in the values: \(I = \frac{0.250 \times 9.8}{2 \times 0.7} = 1.75\text{ A}\).
A long straight wire carries an electric current \(4\text{ A}\). The magnetic induction at a perpendicular distance \(2\text{ m}\) from the wire is
The magnetic field near a long straight wire is given by \(B = \frac{\mu_0 I}{2\pi r}\). Substituting \(I = 4\text{ A}\) and \(r = 2\text{ m}\) with \(\mu_0 = 4\pi \times 10^{-7}\text{ T m/A}\) gives \(B = \frac{4\pi \times 10^{-7} \times 4}{4\pi} = 4 \times 10^{-7}\text{ T}\).
A circular coil of radius \(R\) having current \(I\) is placed in a uniform magnetic field \(B\). If the angle between the area vector of the coil and the magnetic field is \(60^circ\), then the torque on the coil will be:
The torque is given by \(\tau = MBsin\theta\), where \(M = I A = I(\pi R^2)\) and \(\theta = 60^\circ\). Thus, \(tau = I(\pi R^2)Bsin 60^\circ = \frac{\sqrt{3}\pi R^2 I B}{2}\).
A long solenoid having number of turns per unit length 200 carries a current of \(2.5 \text{ A}\), the magnetic field at the end of the solenoid is
The magnetic field at the end of a long solenoid is \(B_{\text{end}} = \frac{1}{2} \mu_0 n I\). Substituting the given values: \(B_{\text{end}} = \frac{1}{2} (4\pi \times 10^{-7}) (200) (2.5) = 3.14 \times 10^{-4} \text{ T}\).
Assertion (A): Parallel current in wires attracts to each other due to magnetic force.
Reason (R): Two electron beams moving parallel to each other repels to each other due to electric force.
Wires with parallel currents attract due to magnetic force, so (A) is true. Two parallel electron beams experience electric repulsion due to like charges, so (R) is true. However, the magnetic force (A) and electric force (R) are distinct phenomena. Thus, (R) does not explain (A).
Assertion (A): Force on a current carrying wire of length \(dvec{l}\) placed in magnetic field \(vec{B}\) is given by \(d\vec{F} = Id\vec{l} \times \vec{B}\).
Reason (R): Net force on a current carrying loop in a non-uniform magnetic field must be non-zero.
The Lorentz force law states \(d\vec{F} = I(d\vec{l} \times \vec{B})\), so (A) is true. For a loop in a uniform field, net force is zero; in a non-uniform field, it is generally non-zero, so (R) is true. However, (R) is a consequence of the force law, not an explanation of the force law itself.