Question

The circuit shown in the figure is used to make a magnetic balance to weigh objects. The massto be measured is hung from the center of the bar, that is in a uniform magnetic field of1.50 ${\rm T}$ ,directed into the plane of the figure. The battery voltage can be adjusted to vary the current in the circuit. The horizontal bar is60.0 ${\rm cm}$ long and is made of extremely light-weight material. It is connected to the battery by thin vertical wires that can support no appreciable tension; all the weight of the suspended massis supported by the magnetic force on the bar. A resistor with= 5.00 ${\Omega}$ is in series with the bar; the resistance of the rest of the circuit is much less than this.

Part A

Which point,aorb, should be the positive terminal of the battery?

Part B

If the maximum terminal voltage of the battery is175 ${\rm V}$ ,what is the greatest massthat this instrument can measure?

Please explain how you arrive to your answers!! I need to UNDERSTAND for my exam coming up. Thank you.

Answer 1

Concepts and reason

The main concept required to solve this question is force experienced by a rod placed in magnetic field.

To identify the polarity of the two terminals of the battery, initially find the direction of the current in the wire. The direction of current is given by the direction of magnetic force experienced by charged particle.

For the mass of the instrument, first find the current flowing in the circuit by using Ohm’s law. Then use the expression of magnetic force experience by a current carrying

conductor placed in magnetic field.

Equate this force with the weight of the body and solve for the mass of the instrument.

Fundamentals

Consider a current carrying conducting rod placed in magnetic field B.

The magnetic force experienced by a rod placed in magnetic field is given by the following formula:

$F = IlB\sin \theta$

Here, $\theta$ is the angle between the current element in the direction of the current flowing in the loop and magnetic field, l is the length of the rod and I is the current flowing through it.

Let a potential difference of V applied across the rod and R be the resistance in the circuit.

The current flowing through the rod is given as,

$I = \frac{V}{R}$

The weight of a mass m is given by following expression:

$W = mg$

Here, g is the acceleration due to gravity.

(A)

The direction of the magnetic force experienced by charged rod is given by the following expression:

$\vec F = I\left( {\vec l \times \vec B} \right)$

The weight of the instrument acts vertically downward. So, to hold the mass, the direction of magnetic force should be vertically upwards.

Since, the direction of magnetic field is inwards hence, the direction of $\vec l$ must be from left to right to make the fore in the upward direction (The resultant vector of cross product of two vectors always points in the perpendicular direction. The direction can be finding out by the following method:

If first figure of right hand will show the direction of the current and the middle figure will show the direction of the magnetic field then the thumb will show the direction of force.). Hence, the direction of current is anticlockwise.

Since current flows from positive to negative terminal hence, the terminal a will be positive and terminal b will be negative.

(B)

Consider a current carrying conducting rod placed in magnetic field B.

The magnetic force experienced by the rod is,

$F = IlB\sin \theta$ …… (1)

The current flowing through the rod is given as,

$I = \frac{V}{R}$

Substitute $\frac{V}{R}$ for I in equation (1).

$F = \frac{V}{R}lB\sin \theta$ …… (2)

The weight of a mass m is given by following expression:

$W = mg$

Since weight will be balanced by the force given in equation (2) hence, substitute mg for F in equation (2).

$mg = \frac{V}{R}lB\sin \theta$
‎

Rearrange the above equation for m.

$m = \frac{V}{{gR}}lB\sin \theta$

Substitute $175{\rm{ V}}$ for V, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g , $5\Omega$ for R , $1.50{\rm{ T}}$ for B , $90^\circ$ for $\theta$ and $60.0{\rm{ cm}}$ for l.

$\begin{array}{c}\\m = \frac{{\left( {175{\rm{ V}}} \right)\left( {60.0{\rm{ cm}}} \right)\left( {1.50{\rm{ T}}} \right)\sin \left( {90^\circ } \right)}}{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {5\Omega } \right)}}\\\\ = \frac{{\left( {175{\rm{ V}}} \right)\left( {60.0{\rm{ cm}}} \right)\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)\left( {1.50{\rm{ T}}} \right)\sin \left( {90^\circ } \right)}}{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {5\Omega } \right)}}\\\\ = 3.21{\rm{ kg}}\\\end{array}$

Ans: Part A

The point a is the positive terminal of the battery.

Part B

The greatest mass that the instrument can measure is $3.21{\rm{ kg}}$ .