A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is :

ir B
2priB
zero
priB

Solution

The force on each point on loop is radially outward and so net force = 0

Electron move at right angle to a magnetic field of 1.5 × 10^–2 tesla with speed of 6 × 10⁷ m/s. If the specific charge of the electron is 1.7 × 10¹¹ C/kg. The radius of circular path will be

3.31 cm
4.31 cm
1.31 cm
2.35 cm

Solution

A charged particle moves through a magnetic field in a direction perpendicular to it. Then the

velocity remains unchanged
speed of the particle remains unchanged
direction of the particle remains unchanged
acceleration remains unchanged

Solution

Magnetic force acts perpendicular to the velocity. Hence speed remains constant.

A current loop in a magnetic field

can be in equilibrium in one orientation
can be in equilibrium in two orientations, both the equilibrium states are unstable
can be in equilibrium in two orientations, one stable while the other is unstable
experiences a torque whether the field is uniform or non-uniform in all orientations

Solution

A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be

2 √3 J
3 J
√3 J
³⁄₂ j

Solution

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an -particle to describe a circle of same radius in the same field?’

2 MeV
1 MeV
0.5 MeV
4 MeV

Solution

An alternating electric field, of frequency v, is applied across the dees (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field(B) used in the cyclotron and the kinetic energy (K) of the proton beam, produced by it, are given by

Solution

Two similar coils of radius Rare lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2 I, respectively. The resultant magnetic field induction at the centre will be

\(\frac{\sqrt{5}\mu _{0}I}{2R}\)
\(\frac{3\mu _{0}I}{2R}\)
\(\frac{\mu _{0}I}{2R}\)
\(\frac{\mu _{0}I}{R}\)

Solution

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency fHz.The magnitude of magnetic induction at the centre of the ring is

\(\frac{\mu _{0}qf}{2R}\)
\(\frac{\mu _{0}q}{2fR}\)
\(\frac{\mu _{0}q}{2\pi fR}\)
\(\frac{\mu _{0}qf}{2\pi R}\)

Solution

Magnetic field at the centre of the ring is \(\frac{\mu _{0}qf}{2R}\)

A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I₁ at a distance d from the conductor as shown in figure. The loop will experience

a net repulsive force away from the conductor
a net torque acting upward perpendicular to the horizontal plane
a net torque acting downward normal to the horizontal plane
a net attractive force towards the conductor

Solution

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