Moving Charges And Magnetism – NEET UG PhysicsPractice Questions & PYQs

The net magnetic field at the center is determined by combining the individual magnetic effects of the circular currents according to their opposing directions. Simplifying the given expression $B = \frac{\mu_{0}}{4 \pi} \frac{2 i_{2}}{r / 2} - \frac{\mu_{0}}{4 \pi} \frac{2 i_{1}}{r / 2}$ yields the combined form $\frac{\mu_{0}}{4 \pi} \frac{4}{r} (i_{2}-i_{1})$. Substituting the values where $i_{2}$ is 2.5, $i_{1}$ is 5.0, and $r$ is 5 completes the evaluation to yield $4 \mu_{0} \pi$, while the negative indicator confirms that the field acts inward into the plane.

Maintaining horizontal motion for a charged particle requires that the downward gravitational force be precisely cancelled by an opposing vertical force supplied by electric or magnetic fields. Statement (3) is viable because mutually perpendicular electric and magnetic fields can both generate vertical forces capable of balancing gravity during horizontal travel. Statement (4) is also correct because a magnetic field running parallel to the particle velocity produces zero magnetic force, leaving a properly oriented perpendicular electric field to solely sustain the vertical equilibrium.

According to Ampère's law, the magnetic field around a symmetric current-carrying conductor is determined entirely by the net current enclosed within an imaginary Amperian loop. Inside a hollow pipe, any chosen circular path encloses zero current, making the internal magnetic field vanish completely. However, an Amperian loop drawn outside the pipe encloses the total current flowing through the conductor, resulting in a non-zero magnetic field in the region exterior to the pipe.

The rotational force experienced by a current-carrying solenoid in a uniform magnetic field is governed by the equation $\tau = M B \sin\theta$, where $M$ is the magnetic dipole moment. First, compute the magnetic moment by multiplying the number of turns, the current, and the cross-sectional area, yielding $M = 2000 \times 2.0 \times 1.5 \times 10^{-4}$, which equals $0.6$ joules per tesla. Next, substitute this dipole moment, the magnetic field strength, and the sine of the angle $30^\circ$ into the torque formula to obtain $\tau = 0.6 \times 5 \times 10^{-2} \times \sin(30^\circ)$, resulting in $1.5 \times 10^{-2}$ Newton-meters.