Magnetic Balance. The circuit shown in Fig. E27.39 is used to make a magnetic balance to weigh objects. The mass m to be measured is hung from the centre of the bar that is in a uniform magnetic field of 1.50 T, directed into the plane of the figure. The battery voltage can be adjusted to vary the current in the circuit. The horizontal bar is 60.0 cmlong 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 mass m is supported by the magnetic force on the bar. A resistor with$\mathbf{R}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\Omega }$is in series with the bar; the resistance of the rest of the circuit is much less than this. (a) Which point, a or b, should be the positive terminal of the battery? (b) If the maximum terminal voltage of the battery is 175 V, what is the greatest mass m that this instrument can measure?

The horizontal bar is, l = 60.0 cm .

The uniform magnetic field is, B = 1.50 T .

The resistor is, $\mathrm{R}=5.00\mathrm{\Omega }$.

The maximum terminal voltage of the battery is, V = 1775 V .

A magnetic field is a field that describes the magnetic influence on a space object.

(a)

By varying current, the battery voltage is adjusted.to measure mass m apply magnetic field which has same magnitude of gravitational force in opposite direction so the magnetic field points upwards.by right hand rule current in the circuit flows in counter clockwise direction.

Thus, for this current flow the positive terminal of battery is a.

(b)

Maximum mass that instrument measure for V = 175 V is,

Angle between magnetic field and length is .magnitude of force is,

$F=llB\sin \varphi$

And

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

So,

$\begin{array}{r}F=\frac{VIB\sin \varphi }{R}\\ F=\frac{VIB}{R}\end{array}$ (1)

Here I is the current, l is the wire length, R is the resistance and B is the magnetic field.

For gravitational force is expressed as,

F = mg …(2)

From the equation (1) and (2).

$\begin{array}{r}mg=\frac{VIB}{R}\\ m=\frac{VIB}{gR}\end{array}$

Here m is the mass and g is the gravitational acceleration.

Substitute all the value in the above equation.

$\begin{array}{r}\mathrm{m}=\frac{\left(175\mathrm{V}\right)(0.600\mathrm{m})(1.50\mathrm{T})}{(5.00\mathrm{\Omega })\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)}\\ =3.214\mathrm{kg}\\ \mathrm{m}=3.21\mathrm{kg}\end{array}$

Thus, the greatest mass m that this instrument can measure is m = 3.21kg .