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Problem 55 Easy Difficulty

A rectangular coil with resistance $R$ has $N$ turns, each of length $\ell$ and width $w,$ as shown in Figure $\mathrm{P} 20.55 .$ The coil moves into a uniform magnetic field $\overrightarrow{\mathbf{B}}$ with constant velocity $\overrightarrow{\mathbf{v}}$ . What are the magnitude and direction of the total magnetic force on the coil (a) as it enters the magnetic field, (b) as it moves within the field, and (c) as it leaves the field?

Answer

a. F=\frac{N^{2} B^{2} w^{2} v}{R}, \text { to the left }
b. zero
c. F=\frac{N^{2} B^{2} w^{2} v}{R}, \text { to the left }

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Video Transcript

hi. In the given problem, there is a magnetic field perpendicular to the plane off paper and directed into it. Then there is rectangular coil, which is having and turns having length l and with W It is entering into this magnetic field which is having a magnet your B with our velocity be so this width off the coil will be intercepting the magnetic field lines perpendicular early hands emotional IMF will be induced across the ants off with. In first part of the problem, emotional e M f will be induced across the ends off direct, angular loop or soil which will be given us the product off magnetic field with the velocity and with the length of this arm, which is W now as the resistance off this oil is our so the current passing through it is given using homes law as e m f divided by our resistance. So this has given us bbw by, uh now, as this width is a current carrying conductor And yeah, of course, the soil is having and turns Sonett e m f induced a cruise. The ends of the earth will be n b we W So this is N B we w by are the current passing through this coil? No, the this current carrying arm this with having the word www with arm having the will to w as it is moving in the magnetic field. So it will experience a magnetic force which is given by the product off magnetic field and current and the resistance and the length off the arm which is again w So it becomes for be it means same for I This is and be we w by our in tow w So finally the expression for this force becomes and end times Yes, as there are total en arms having WCW so the total force will be end times off this single force. So this is any square Be square we into w squared divided by our now in order to find the direction of this force as the coil is moving towards the right. So as per lenses law toe oppose this rightward motion. This force should be acting toe words left Asper Lenses Law Now in second part of the problem, then the loop is moving inside the magnetic field inside the magnetic field, both off its arms. Having doubled, W will be moving in the same magnetic field with the same velocity. So both of them will be having the IMF induced in the same direction. Hence these two in EFS will neutralize each other so there will be no net enough induced across the coil or we can say no net current. So no net force will be acting on the oil or Luke. Now, in the third part of the problem, when this coil is coming out off this magnetic field from here, it will be coming out with the same velocity. So again, this arm, the opposite arm having the same with the W, it will be intercepting the magnetic field lines for particularly so same e m f will be induced across its ends given by nb VW. So same current will be passing through the coil. So same force will be experienced by this coil and in the same direction means towards right towards left. Same force will be acting on the loop and in the same direction means left the word. So this is the answer for the third part off that problem. Thank you

H.B.T.I.
Top Physics 102 Electricity and Magnetism Educators
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