The prototype shows that magnetic field cannot be kept constant in a wide range of 3 m. Therefore you reduce the length of the bar to 0.2 m keeping all other remaining parameter unchanged. What is the power exerted?

1 W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/25 W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/9 W

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 W

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 1/9 W
$P = \frac{(B l v)^{2}}{0.9}$

$= \frac{(0.25 \times 0.2 \times 2)^{2}}{0.9} = \frac{(.1)^{2}}{.09}$

$= 1 / 9 W$

Suggest Corrections

Similar questions

Q. A cathode of photoelectric cell is changed such that the work function changes from

W_{1}

W_{2}

(

W_{1} < W_{2}

). If the current before and after changes are

I_{1}

and

I_{2}

all other conditions remaining unchanged, then (assuming

h v > W_{2}

The cathode of a photoelectric cell is changed such that the work function changes from $W_{1} t o W_{2} (W_{2} > W_{1})$ . If the current before and after change are $l_{1} a n d l_{2}$ all other conditions remaining unchanged, then (assuming hn > $W_{2}$ )

Q. An elastic bar of length

L

, uniform cross sectional area

A

, coefficient of thermal expansion

α

and Young's modulus

E

is fixed at the two ends. The temperature of the bar is increased by

T

, resulting in an axial stress

σ

. Keeping all other parameters unchanged, if the length of the bar is doubled, the axial stress would be

A bar magnet of length 0.2 m and pole strength 5 A.m. should be kept in a uniform magnetic field of induction 15 tesla at angle ..... radians to the field so that the torque experienced by it will be 7.5N-m

Join BYJU'S Learning Program

The prototype shows that magnetic field cannot be kept constant in a wide range of 3 m. Therefore you reduce the length of the bar to 0.2 m keeping all other remaining parameter unchanged. What is the power exerted?

The correct option is D 1/9 WP=(Blv)20.9=(0.25×0.2×2)20.9=(.1)2.09=1/9W

The correct option is D 1/9 W
$P = \frac{(B l v)^{2}}{0.9}$

$= \frac{(0.25 \times 0.2 \times 2)^{2}}{0.9} = \frac{(.1)^{2}}{.09}$

$= 1 / 9 W$