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Chapter 4: Q39E (page 984)

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the centre of the solenoid and 3.50 cm from its axis is 8.00 x 10-6 V/m. Calculate di/dt

Short Answer

Expert verified

If a long thin solenoid having 400 turns per meter and radius of , the current in the solenoid is increasing at a unform rate of 9.21A/s

Step by step solution

01

Calculating the value of E

It is given that a long solenoid of 400 turns and distance of m-1and a radius of 1.10cm . The induced electric field at a point near the centre of the solenoid is 3.50cm and induced electric field is 8.00脳10-6V/m

Now using Faraday鈥檚 law we get:

蔚=鈭�E鈫�.dl鈫�=-d蠒Bdt

Now, the magnetic flux through the circle of radius r > R :

蠒=AB=蟺搁2B

Now integrating LHS we get:

鈭�E鈫�.dl鈫�=E2蟺r

Absolute value of E is:

E2蟺r=蟺R2dBdt

We can also write it as:

E=R2渭0n2r=dBdt

02

Calculating the values

Now to get dldtwe get:

dldt=2Er渭0nR2

Now, substituting the values we get:

28.00脳10-6V/m0.0350m4蟺脳10-7T.m/A4000.0110m2=9.21A/s

Therefore, the value ofdldt is = 9.21A/s

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Most popular questions from this chapter

A long, straight solenoid with a cross-sectional area of 8.00 cm2 is wound with 90 turns of wire per centimetre, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its centre. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) and (b) the terminal voltage Vabof the 16.0-V battery.

Fig. E25.33

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