Consider $$d >> a$$, 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)
$$T = MB \sin \theta = I \pi a^2 \times \displaystyle \dfrac{\mu_0 I}{ \pi d} \sin 30^{\circ} = \dfrac{\mu_0 I^2 a^2}{2d}$$
At some location on earth the horizontal components of earth's magnetic field is $$18 \times 10^-6 T.$$ At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes $$45^0$$ angle with horizontal in equilibrium to keep this needle horizontal, the vertical force that should be applied at one of its ends is:
A charge $$q$$ is spread uniformly over an insulated loop of radius $$r$$. If it is rotated with an angular velocity $$\omega$$ with respect to normal axis then the magnetic moment of the loop is
If M is magnetic moment and B is the magnetic field, then the torque is given by:
A closely wound solenoid of 2000 turns and area of cross-section $$1.5 \times 10^{-4} m^2$$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $$5 \times 10^{-2}$$ tesla making an angle of 30$$^o$$ with the axis of the solenoid. The torque on the solenoid will be
A current loop in a magnetic field
A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength $$0.2 \ Weber / m^2$$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $$30^o$$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be
A closely wound solenoid of 2000 turns and area of cross-section $$1.5 \times 10^{-4}m^2$$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $$5 \times 10^{-2}T$$, making an angle of 30$$^o$$ with the axis of the solenoid. The torque on the solenoid will be
A magnetic needle lying parallel to the magnetic field required W units of work to turn it through an angle 45$$^o$$. The torque required to maintain the needle in this position will be
If a solenoid of magnetic moment $$0.2Am^2$$ in a uniform magnetic field $$0.16$$T experiences a torque $$0.016$$Nm, then angle made by a solenoid axis with magnetic field is _____________.
The magnetic field in the cylindrical region shown in figure increase at a constant rate of $$20.0mT/s$$. Each side of the square loop $$abcd$$ and $$defa$$ has a length of $$1.00cm$$ and a resistance of $$4.00\Omega$$. Find the current (magnitude and sense) in the wire $$ad$$ if (a) the switch $${S}_{1}$$ is closed but $${S}_{2}$$ is opne, (b) $${S}_{1}$$ is open but $${S}_{2}$$ is closed, (c) both $${S}_{1}$$ and $${S}_{2}$$ are open and (d) both $${S}_{1}$$ and $${S}_{2}$$ are closed.