Question

56. A conducting rod (AB) of length 30 cm, having
resistance 0.5 g is moving on two parallel
horizontal, conducting and resistanceless rail as
shown in the figure. Current in resistance R=2.5
60
Q resistor is
B=021
R=250
→v=20 mis
(1) 0.1 A
(3) 0.3 A
(2) 02A
(4) 04A​

Answer 1

Answer:

The current in the resistor will be 0.4 Amperes.

Given,

length = 30 cm

Resistance of rod(r) = 0.5 Ω

Resistance of current(R) = 2.5 Ω

On conversion of the length into meters we get,

length = 30 * 10⁻² m

The magnetic field is given to be 0.2 T and its hundred times is V which comes out to be 20 m/s

Here, we know

The magnetic field is always accompanied by the electric field.

With the help of the formula of the magnetic field, we can identify the current.

The formula we can use is:

I = BLV

= 0.2 T * 30 * 10⁻² m * 20

= 120 * 10⁻² V

We also know that,

i = v / (r + R)

i = (120 * 10⁻² ) / (0.5 + 2.5)

i = (120 * 10⁻² ) / (3)

i = 0.4 A

Therefore, the current in the resistance R = 2.5 Ω is 0.4 A.

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Answer 2

Answer: The current in the resistor will be 0.4 Amperes.

Given the length of a conducting rod (30 cm), its resistance (0.5 ohms), and the resistance of a resistor (2.5 ohms), we can determine the current flowing through the resistor.

We know that the magnetic field accompanying the electric field is 0.2 T, which equates to 20 m/s when multiplied by 100. Using the formula for the magnetic field (I = BLV), we can calculate the current.

By converting the length of the rod to meters (30 cm = 30 x 10^-2 m), we can find the current as

                        0.2 T * 30 x 10^-2 m * 20 = 120 x 10^-2 V.

Using Ohm's law (i = v / (r + R)), we can calculate the current flowing through the resistor as

                      i = (120 x 10^-2 ) / (0.5 + 2.5) = (120 x 10^-2 ) / 3 = 0.4 A.

Therefore, the current in the resistor is 0.4 Amperes.

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