A straight wire of mass 10.0 g and length 5.0 cm is sus- pended from two identical springs that, in turn, form a closed circuit (Fig. P19.74). The springs stretch a dis- tance of 0.50 cm under the weight of the wire. The cir- cuit has a total resistance of 12 0. When a magnetic field directed out of the page (indicated by the dots in the fig- ure) is turned on, the springs are observed to stretch an additional 0.30 cm. What is the strength of the magnetic field? (The upper portion of the circuit is fixed.) 24 V MM 5.0 cm

The required strength of the magnetic field is 0.59 T

Step-by-step explanation

Given,

mass of wire, m =10.0 g =10*10^-3 kg

length ,L= 5.0 cm =5*10^-2 m

resistance, R =12

Voltage, V = 24

Let x₁ be the elongation in the spring system in the absence of the magnetic field and x₂ be the additional stretch in the spring system in the presence of magnetic field.

x₁ = 0.50 cm

x₂ = 0.30 cm

Now at equilibrium,

F_r = F_g + F_m, where

F_r restoring force in the absence of magnetic field = kx₁

so, F_r' = k(x₁ + x₂) ie the restoring force in the presence of magnetic field

F_g = mg = gravitational force

F_m = magnetic force

so,

k(x₁ + x₂) = mg +F_m

F_m = mg - k(x₁ + x₂)

F_m = [mg/x₁] (x₁ + x₂) - mg

= mg { [ x₁ + x₂/x₁ ] -1 }

F_m = mg {x₂ / x₁ }

According to Ohm's Law,

V = IR

I = V/R

F_m = BIL sin?θ?

Here, ?θ? is the angle between the direction of current and direction of magnetic field.The direction of current is perpendicular to the magnetic field vector.so,

F_m = BIL sin 90?°? = BIL

B = F_m / IL

Substituting the values of F_m and I, we get,

B = [mg{x₂/x₁} ] / {V/R}L

B = mgx₂R / VLx₁

B =[ {10*10^-3}*9.81*{0.30*10^-2}*12 ] / [ 24*{5*10^-2}*{0.50*10^-2}]

B =0.59T

Therefore the strength of the field is 0.59T