4. A long and narrow rectangular loop of wire is moving...

4. A long and narrow rectangular loop of wire is moving toward the bottom of the page with a speed of 0.05 ms?¹. The loop is leaving a region in which a 2.4T magnetic field exists; the magnetic field outside this region is zero. During a time of 3.0 s, what is the magnitude of the change in the magnetic flux? 0.08m 5.An air core solenoid has 300 turns, length 20cm and cross-sectional area of 4 cm². A current flows in the coil at a rate of 50 As?¹. Calculate the emf induced in the solenoid. How much energy is stored in the solenoid? 6.Suppose you wish to make a solenoid whose self-inductance is 1.4 mH. The inductor is to have a cross-sectional area of 1.2 x 10?³ m² and a length of 0.052m. How many turns of wire are needed?

Transcript

00:01 Hello students in the fourth question we are given that a long narrow rectangular loop of wires moving towards the bottom of the page with a speed of here 0 .05 meters per second and the loop is leaving a region in which a 2 .4 tesla magnetic field exists.

00:21 So b is 2 .4 tesla and the magnetic field outside this reason is zero.

00:27 So during a time of 3 seconds.

00:30 What is the magnitude of the change in magnetic flux now if here let the length of rectangular loop length of rectangular loop is l and the breadth is given to us which is equal to 0 .08 meters here this is 0 .08 meters here this is 0 .08 meters so let us see figure so where this is the figure now the flux we have we know that the magnetic flux is given by b a b multiply cost theta the initial flux or we can say flush flux when it is in the magnetic field is equal to magnetic field is 2 .4 multiply by area length multiply by breadth it is 0 .08 multiplied by l or we can write it as here length multiply by breadth which is 0 .08 multiply by cos 0 degree.

01:49 So from here the initial flux comes out to be 0 .192 l weber and after 3 seconds here after 3 seconds the flux is equal to the magnetic field is 0 so after the movement of 3 seconds it will be here b is 2 .4 multiply l is areas l multiply by 0 .08 over here and we have or the length will change breath will remain same we can write it as 0 .08 and length will be here speed is 0 .05 meters per second so after in three seconds it will move vt distance so length will be decreased by vt so l minus 0 .05 multiplied by 3 so from here we have 5 f is equal to this is equal to 0 .088 here so here or we can write it as 5 f is equal to 2 .4 multiply by 0 .08 multiplied by l minus 2 .4 multiplied by 0 .08 multiplied by 0 .05 multiplied by 3 .5 will be so this is here this will be 0192l 0 .192l minus it this is 0 .0 to 88 so change in flux will be 5 minus 5 i say it is 0 .192 l minus 0 .088 minus 5i minus 5i which is 0 .192l minus 5i which is 0 .192 l so from here we have the change in flux will be equal to this is here minus 0 .28 minus 0 .28 -028 -2 -8 -8 - if we take the magnitude, so it will be minus 0 .08 -8 -weber.

04:12 Now in the next question here in the next question the fourth question in the or in the here this is fifth question we are given that an air core solenoid has 300 tons so the value of n we have been given so the value of n is equal to 300 tons over air and the length of the solenoid is is equal to 20 cm which is equal to 20 multiplied 10 to 2 meters and the cross -sectional area is equal to 4 cm square which is 4 multiplied 10 to 2 2b minus 4 meter square so the current flows in the coil at a rate of 50 amp per second so we are given the i over d t which is 50 amp per second so we have to find the amf induced in the solenoid and how much energy is stored in the solenoid.

05:20 So here we know that emf induced is equal to n d phi over d t and we know that phi is equal to b multiplied by a and magnetic will do to solenoid is mu not n i.

05:40 So from here we have e is equal to n d over d t of b.

05:47 B, a.

05:49 So from here we have e is equal to n.

05:52 A is constant area due over dt of b.

05:57 So b is mu not n i over here...

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