Question

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Here is a question asked in JEE Advanced 2003.

Chapter: Magnetics
Type: Subjective.


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Answer 1

The axle of a circular wheel of radius R is held horizontally by two identical strings of equal lengths separated by a distance D. The tension in each string is

T

0

. The rim of the wheel carries a total charge +Q distributed uniformly on it. The wheel is vertical and is kept in a uniform vertical magnetic field

B

. It is now rotated at an angular speed ω. If the string break at a tension of 3T

0

/2, than the maximum possible value of ωat which the wheel can be rotated without breaking a string is QBR

2DTO.

For the situation described in figure , the magnetic field changes with time according to

B=(2.00t3-4.00t2+0.8)T

and

r2=2R=5.0cm

a) Calculate the force on an electron located at

P2

at t=2.00 s

b) What are the magnitude and direction of the electric field at

P1

when t=3.00 s and

r1=0.02m

Explanation:

I HOPE IT HELPS.

Answer 2

✪ANSWER:-

➪$\omega = dTo/ QR {}^{2} b$

✪SOLUTION :-

》 2To = mg

》 I = Q / T

➪$》 \frac{q}{2\pi \div \omega} = \frac{q \omega }{2\pi}$

=> Magnetic moment of loop is ,,

M = i A

=> Now torque is ,

T = MB sin90°

If T1 & T2 are tensions in the string,

➪ (T1 - T2 ) d / 2 = i AB

➪ T1 - T2 = 2iAB / d ------(1)

➪ T1 + T2 = mg --------(2)

From eqns 1 & 2,

➪ T1 = mg/2 + (iAB/d )

and , ➪ T2 = mg/2 - (iAB/d)

We know ,T1 > T2

➪Hence, T1 = 3To/ 2

Hence,

➪ 3To/ 2 = 2To /2 +$\omega$ Q / 2 . pi. d × pi R^2 × B

➪ To = $\omega$ QBR^2 / d

HENCE,

➪ $\omega$= dTo / QR^2 B

HOPE IT HELPS :)