Aclosely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 o. it is connected to a charge-measuring instrument whose resistance is 45.5 o. when the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrument indicates a charge of 3.53 x 10-5 c. what is the magnitude of the magnetic field?

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

Where,

electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

And that the current can be expressed as function of charge and time, then

Equation Faraday's law and Ohm's law we have,

Re-arrange for Magnetic Field B, we have

Our values are given as,

Replacing,

Therefore the magnetic field in the System is 0.095T

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards