JEE Main 2017 (Online) 8th April Morning Slot | Magnetics Question 222 | Physics

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JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

A magnetic dipole in a constant magnetic field has :

maximum potential energy when the torque is maximum.

zero potential energy when the torque is minimum.

zero potential energy when the torque is maximum.

minimum potential energy when the torque is maximum.

JEE Main 2017 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

Magnetic field in a plane electromagnetic wave is given by

$$\overrightarrow B $$ = B₀ sin (k x + $$\omega $$t) $$\widehat j\,T$$

Expression for corresponding electric field will be :
Where c is speed of light.

$$\overrightarrow E $$ = B₀ c sin (k x + $$\omega $$t) $$\widehat k$$ V/m

$$\overrightarrow E $$ = $${{{B_0}} \over c}$$ sin (k x + $$\omega $$t) $$\widehat k$$ V/m

$$\overrightarrow E $$ = $$-$$ B₀ c sin (kx +$$\omega $$t) $$\widehat k$$ V/m

$$\overrightarrow E $$ = B₀ c sin (kx $$-$$$$\omega $$t) $$\widehat k$$ V/m

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

A magnetic needle of magnetic moment 6.7 $$\times$$ 10^-2 A m² and moment of inertia 7.5 $$\times$$ 10^-6 kg m² is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

8.76 s

6.65 s

8.89 s

6.98 s

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$$\Omega $$, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10V is:

4.005 × 10³ $$\Omega $$

1.985 × 10³ $$\Omega $$

2.535 × 10³ $$\Omega $$

2.045 × 10³ $$\Omega $$

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