A 50-turn circular coil (radius r = 15 cm) with a total...

A 50-turn circular coil (radius r = 15 cm) with a total resistance R = 4.0 ? is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of this field varies with time as B = A sin(?t), where A = 80 ?T and ? = 50? rad/s. a) Find the electric flux through the coils at time t = 20 ms. b) Find the induced emf in the coils at time t = 20 ms. c) What is the magnitude of the current induced in the coil at time t = 20 ms?

Transcript

00:01 Hello students in this question we are given the number of terms in the coil which is here 50 and we are given the radius of the coil which is 15 centimeter which is 15 multiplied by 10 to the per minus 2 meters and their total resistance in the coil is 4 .0 oms which is placed in a magnetic field directed perpendicular to the plane of the coil and the magnet the magnetic field varies as a sine omega t where the value of a is 80 micro tesla which is 8t multiplied by 10 to the minus 6 tesla and where omega is equal to 50 pi radian per second so in the a part we have to find out what we have to find the electric flux through the coil at time t equal to 20 milliseconds so we know that flux is b dot a.

01:08 So we have 5 is equal to b.

01:13 So b is 80 multiplied by 10 to the 0 .6.

01:18 Sign 50 pi t.

01:23 So here, sign 55 t is 20 milliseconds 20 multiplied by 10 to the plus minus 3 multiplied by it is area.

01:34 So it will be pi r squared by pi multiplied by r that is 15 here this is 15 multipripe by 10 to the par minus 2 whole square so here five comes out to be so here it comes out to be 80 multiply 10 to the per minus 6 multipri by pi multiply by 15 multiply by 10 to the per minus 2 whole square sign we have 505 multi 5 multiply by 10 to the per minus 2 whole square sign we have 50 pi multi multiplied by 20 multiplied by 10 to the minus 3 so this will be sine pi now we know that sine pi is 0 so the whole term will be 0 so the electric flux is 0 newton coulum meter square or we can write here electric flux is 0 newton meters square per coolum now in the b part we have to find what is the induced current in the coil at time t equal to 20 milliseconds now we know that e is equal to minus d5 over dt.

02:45 So here we have minus d or we can write it as magnitude.

02:55 So the minus sign will be removed over here.

02:58 So it is d over dt of a sine omega t multiplied by area a naught...

Question

Please give Ace some feedback