NAEST 2020: Magnet placed in the vicinity of a current carry
By Question: When a current is passed through the coil wrapped on the syringe, a force of attraction or repulsion is exerted on the magnet which goes towards or away from the coil. The force acts on the magnet because
- The coil produces a non-uniform magnetic field near itself
- The coil produces a uniform magnetic field near itself
- The magnet produces a non-uniform magnetic field near itself
- The magnet produces a uniform magnetic field near itself
Question 2: An induced electric field was produced in the experiments in this video in the period
- when the battery was connected and the current in the coil increased from zero to the final value.
- when the magnet was moving away from the coil
- when the magnet was moving towards the coil
- when the magnet was staying at the centre of the coil
- No induced electric field was produced in the video.
Solution 1: The magnet may be considered as a magnetic dipole. The dipole consists of a north pole of strength $m$ and a south pole strength $m$, and these poles are separated by a small distance $d$. When placed in a magnetic field $B$, a north pole of strength $m$ experiences a force $mB$ in the direction of the field and a south pole of strength $m$ experiences a force $mB$ opposite to the direction of the field.
A uniform field has the same magnitude and direction at all points in space. If the coil produces a uniform field (in the region of the magnet) then the two poles will experience equal and opposite force. The net force on the magnet will be zero. However, the magnet in this demo experiences a force as it starts moving from the rest. This is possible only when the coil produces a non-uniform magnetic field. Thus, option (A) is correct.
Another way to analyze the problem is from the perspective of the coil. A current-carrying coil is placed in a magnetic field. This current-carrying coil experiences a magnetic force given by $i\vec{l}\times\vec{B}$. Let us consider one circular turn of the coil. Suppose the magnet produces a uniform magnetic field (perpendicular to the plane of the circular turn). The force acting on each element of the circular turn is radial. The net force on the circular turn is zero. This is true for all turns of the coil. Hence, the net force on the coil will be zero if the magnet produces a uniform magnetic field. However, since the magnet experiences a non-zero force, the coil should also experience a non-zero magnetic force (similar to Newton's third law). This is possible only when the magnet produces a non-uniform magnetic field. Thus, option (C) is correct.
Solution 2: The induced current (or emf) is produced when magnetic flux through the coil varies with time. The flux through the coil is the sum total of the flux through each turn of the coil. The flux through a turn depends on the magnetic field in the region of the turn ($\phi=\oint\vec{B}\cdot d\vec{S}$).
When the battery is connected, a current passes through the coil. This current-carrying coil produces a magnetic field. This field results in flux through the coil. As the current increases, the magnetic field increases, and hence flux through the coil also increases. This time-varying flux induces an emf and induced current is set up in the coil. This is known as self induction (setting of induced current in the coil due to time-varying current in it). Thus, option (A) is correct.
Let us consider the situation when the current through the coil reaches its maximum value. The flux through the coil becomes constant and the induced current due to self induction becomes zero. What about the flux through the coil as the magnet moves? The magnetic field of the magnet results in flux through the coil. If the magnet is fixed then this flux is constant. As the magnet moves, the flux through the coil changes due to a change in magnetic field at the location of the coil. This time-varying flux induces emf and induced current in the coil. Thus, options (B) and (C) are correct.
A time-varying magnetic field always produce an induced electric field. Hence, option (E) is not correct.
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